# Trying to Find a Way to Solve: int_0^{infinity} (e^{-st-((a^2)/(4t))}) / (sqrt{pi*t}) dt

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#### mario99

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[imath]\\\\\\[/imath]

[imath]\displaystyle \int_{0}^{\infty}\frac{e^{-st - \frac{a^2}{4t}}}{\sqrt{\pi t}} \ dt = \frac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

Is there a way to solve this integral to get that result?

[imath]\\\\\\[/imath]

[imath]\displaystyle \int_{0}^{\infty}\frac{e^{-st - \frac{a^2}{4t}}}{\sqrt{\pi t}} \ dt = \frac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

Is there a way to solve this integral to get that result?
Did you try substitution:

$$\displaystyle \sqrt{t} = x$$

[imath]\\\\\\[/imath]

[imath]\displaystyle \int_{0}^{\infty}\frac{e^{-st - \frac{a^2}{4t}}}{\sqrt{\pi t}} \ dt = \frac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

Is there a way to solve this integral to get that result?
I haven't looked it up, but if that's what a table says, then it's likely to be true.

What have you tried?

-Dan

Did you try substitution:

$$\displaystyle \sqrt{t} = x$$
When [imath]\displaystyle x = \sqrt{t}[/imath], the integral becomes

[imath]\displaystyle \frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-sx^2-\frac{a^2}{4x^2}} \ dx[/imath]

I haven't looked it up, but if that's what a table says, then it's likely to be true.

What have you tried?

-Dan
I have this idea

[imath]\displaystyle -sx^2 + a\sqrt{s} - a\sqrt{s} - \frac{a^2}{4x^2} = -sx^2 - \frac{a^2}{4x^2}[/imath]

And

[imath]\displaystyle -sx^2 + a\sqrt{s} - \frac{a^2}{4x^2} = -\left(\sqrt{s}x - \frac{a}{2x}\right)^2[/imath]

Then

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-\left(\sqrt{s}x - \frac{a}{2x}\right)^2} \ dx[/imath]

If we choose

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle du = \sqrt{s} + \frac{a}{2x^2} \ \ dx[/imath]

It will not work as we can't isolate [imath]\displaystyle x[/imath].

According to Wolfram Alpha

[imath]\displaystyle x = \frac{\sqrt{a}}{\sqrt{2}\sqrt[4]{s}} \ \ \ [/imath] and [imath]\ \ \ \displaystyle x = -\frac{\sqrt{a}}{\sqrt{2}\sqrt[4]{s}}[/imath]

This is the first time to know that I can solve for two values in solving integrals (assuming they are roots).

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-\left(\sqrt{s}x - \frac{a}{2x}\right)^2} \ dx = \displaystyle \frac{1}{\sqrt{\pi}\sqrt{s}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}e^{-u^2} \ du[/imath]

[imath]= \displaystyle \frac{1}{2\sqrt{s}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{2e^{-u^2}}{\sqrt{\pi}} \ du = \displaystyle \frac{1}{2\sqrt{s}}e^{-a\sqrt{s}} \text{erf}(u)\bigg|_{-\infty}^{\infty} = \displaystyle \frac{1}{2\sqrt{s}}e^{-a\sqrt{s}} (1 + 1) = \displaystyle \frac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

Thank you for the help khansaheb and topsquark. The problem has been solved because of [imath]\displaystyle x = \sqrt{t}[/imath].

I am just wondering how did it work when Wolfram Alpha solve for the roots of [imath]x[/imath].

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According to Wolfram Alpha

[imath]\displaystyle x = \frac{\sqrt{a}}{\sqrt{2}\sqrt[4]{s}} \ \ \ [/imath] and [imath]\ \ \ \displaystyle x = -\frac{\sqrt{a}}{\sqrt{2}\sqrt[4]{s}}[/imath]

This is the first time to know that I can solve for two values in solving integrals (assuming they are roots).

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-\left(\sqrt{s}x - \frac{a}{2x}\right)^2} \ dx = \displaystyle \frac{1}{\sqrt{\pi}\sqrt{s}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}e^{-u^2} \ du[/imath]

[imath]= \displaystyle \frac{1}{2\sqrt{s}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{2e^{-u^2}}{\sqrt{\pi}} \ du = \displaystyle \frac{1}{2\sqrt{s}}e^{-a\sqrt{s}} \text{erf}(u)\bigg|_{-\infty}^{\infty} = \displaystyle \frac{1}{2\sqrt{s}}e^{-a\sqrt{s}} (1 + 1) = \displaystyle \frac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

Thank you for the help khansaheb and topsquark. The problem has been solved because of [imath]\displaystyle x = \sqrt{t}[/imath].

I am just wondering how did it work when Wolfram Alpha solve for the roots of [imath]x[/imath].
Are you saying that you don't know how to integrate
[imath]\int_0^{\infty} e^{-x^2+bx} \, dx[/imath]

That's more or less what this is. (This is another one of those things you should have seen how to do a long time ago.)

You complete the square:
[imath]-x^2 + bx = - ( x^2 - bx )[/imath]

[imath]= - \left ( x^2 - bx + \left ( \dfrac{b^2} {4} - \dfrac{b^2}{4} \right ) \right )[/imath]

[imath]= - \left ( x^2 - bx + \dfrac{b^2} {4} \right ) + \dfrac{b^2}{4}[/imath]

[imath]= - \left ( x - \dfrac{b}{2} \right )^2 + \dfrac{b^2}{4}[/imath]

Then you let u = x - b/2 and take the exponential of [imath]b^2/4[/imath] outside the integration, as it's just a constant.

For the umpteenth time, I need to tell you need to go back and review. You are missing a lot of basics.

-Dan

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle du = \sqrt{s} \, \red{\textbf{dx}}+ \frac{a}{2x^2} \ \ dx[/imath]
C'mon! Partial differential equations and you can't take a simple diffferential?

-Dan

Are you saying that you don't know how to integrate
[imath]\int_0^{\infty} e^{-x^2+bx} \, dx[/imath]

That's more or less what this is. (This is another one of those things you should have seen how to do a long time ago.)

You complete the square:
[imath]-x^2 + bx = - ( x^2 - bx )[/imath]

[imath]= - \left ( x^2 - bx + \left ( \dfrac{b^2} {4} - \dfrac{b^2}{4} \right ) \right )[/imath]

[imath]= - \left ( x^2 - bx + \dfrac{b^2} {4} \right ) + \dfrac{b^2}{4}[/imath]

[imath]= - \left ( x - \dfrac{b}{2} \right )^2 + \dfrac{b^2}{4}[/imath]

Then you let u = x - b/2 and take the exponential of [imath]b^2/4[/imath] outside the integration, as it's just a constant.

For the umpteenth time, I need to tell you need to go back and review. You are missing a lot of basics.

-Dan

C'mon! Partial differential equations and you can't take a simple diffferential?

-Dan
Come on Dan. This is a simple derivative.

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle \frac{du}{dx} = \sqrt{s} + \frac{a}{2x^2} \ \ [/imath]

My question was how to find [imath]x[/imath] when [imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

Can you isolate [imath]x[/imath]?

Wolfram Alpha simply made [imath]\displaystyle u = 0[/imath]

Solved for [imath]x[/imath]
[imath]\displaystyle 0 = \sqrt{s}x - \frac{a}{2x}[/imath]

I said that this is the first time I know we can do this in integrals. We have never found the roots when we made a substitution to solve an integral.

Come on Dan. This is a simple derivative.

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle \frac{du}{dx} = \sqrt{s} + \frac{a}{2x^2} \ \ [/imath]

My question was how to find [imath]x[/imath] when [imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

Can you isolate [imath]x[/imath]?

Wolfram Alpha simply made [imath]\displaystyle u = 0[/imath]

Solved for [imath]x[/imath]
[imath]\displaystyle 0 = \sqrt{s}x - \frac{a}{2x}[/imath]

I said that this is the first time I know we can do this in integrals. We have never found the roots when we made a substitution to solve an integral.
I responded to your line where you took the differential, not the derivative. You left the dx off. It was a silly mistake made by someone going too fast and I would treat any other student in the same way.

Are you actually telling me that you don't know how to solve
[imath]\displaystyle u = \sqrt{s}x - \dfrac{a}{2x}[/imath]

for x? Seriously?? What happened to the guy who solved the Airy equation?

If you really need the hint: Multiply both sides by x. What does the equation look like?

But you did ask about that part. You didn't know where the "factoring" took place and why it worked. I showed you what you needed to do to do the integration. You complete the square and the rest follows. It's a standard method that you should be well-acquainted with at your "level." If you did the work, you would see that and you wouldn't need to use W|A.

If you are wondering why my tone here and in your last thread are different consider that, once again in this thread and after being told an incredible number of times not to, you have posted a question without showing any work. What you finally did post was confused and relied on something that W|A told you and you said you didn't know that you could do the problem that way. My response was to point out corrections to what you wrote and tell you where the integration method comes from. Which is a method that anyone who has progressed to the level of Mathematics you claim to be at would already know how to do. Which means that, yet again, I have to tell you that you need to go back and review instead of moving forward, which I'm tired of telling you and something you don't like hearing.

What, exactly, are you objecting to? If you aren't going to listen to what you are being told to do, and aren't going to do what you are being advised to do, then you have put this on yourself. So, if you want me to treat you like a really good Mathematics student that is fully capable of doing partial differential equations then you have to actually be that student. You did okay in that other thread: I admit that I was impressed; you haven't shown any of that in this thread.

So, do you need help doing that integral or not? I'm willing to help, but you have to show me what you did, not W|A.

-Dan

I responded to your line where you took the differential, not the derivative. You left the dx off. It was a silly mistake made by someone going too fast and I would treat any other student in the same way.
I didn't left the [imath]\displaystyle dx[/imath]. It is a standard way to write [imath]\displaystyle du = \left(\sqrt{s} + \frac{a}{2x^2}\right) \ dx \ \ [/imath] as [imath] \ \ \displaystyle du = \sqrt{s} + \frac{a}{2x^2} \ dx[/imath] in solving such problems. If you paid more attention, you wouldn't need to think the [imath]\displaystyle dx[/imath] was missing in the first term.

Many Calculus books' authors as well as my teacher said that it is OK to write the differentials or the integrals with multiple terms without brackets. In the first case, it is understandable the differential element on the left is for all the terms on the left while the differential element on the right is for all the terms on the right. In integral case, it will look like this

[imath]\displaystyle \int_{1}^{2} x + \cos x \ dx = \int_{1}^{2} (x + \cos x) \ dx[/imath]

Now, after seeing your confusion, I would disagree with these authors and my teacher and I am going to advise to write brackets whenever you have more terms.

But you did ask about that part. You didn't know where the "factoring" took place and why it worked. I showed you what you needed to do to do the integration. You complete the square and the rest follows. It's a standard method that you should be well-acquainted with at your "level." If you did the work, you would see that and you wouldn't need to use W|A.
Now, I understand why you are angry and why you thought I don't know how do completing the square. And I also now understand why you have given me an example to show the method (thank you). You are saying if I know how to do that why I haven't shown it until the hint was given. First, I am used to do completing the square on equations that looks like [imath]\displaystyle ax^2 + bx[/imath] not [imath]\displaystyle ax + \frac{b}{x}[/imath], so I missed that it can be done on the latter as well. Also I was focusing a lot on choosing the correct substitution to get rid of the square root in the denominator. Therefore, I was distracted by these two things. Even experts like you may sometimes get distracted to recognize a simple idea in solving something.

So, do you need help doing that integral or not? I'm willing to help, but you have to show me what you did, not W|A.
Yes.

Are you actually telling me that you don't know how to solve
[imath]\displaystyle u = \sqrt{s}x - \dfrac{a}{2x}[/imath]

for x? Seriously?? What happened to the guy who solved the Airy equation?

If you really need the hint: Multiply both sides by x. What does the equation look like?
I think that you just scanned the problem and didn't actually solve it for [imath]\displaystyle u[/imath], this is why you recommend multiplying both sides by [imath]\displaystyle x[/imath]. I am sorry that I have skipped many steps which I thought that we reached to the level where we can see everything between the gaps. It seems that I have to show you all the steps, so that you can understand why multiplying both sides of the equation by [imath]\displaystyle x[/imath], will not work.

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-\left(\sqrt{s}x - \frac{a}{2x}\right)^2} \ dx[/imath]

If we choose

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle du = \sqrt{s} + \frac{a}{2x^2} \ \ dx[/imath]

[imath]\displaystyle dx = \frac{1}{\sqrt{s} + \frac{a}{2x^2}} \ du[/imath]

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-u^2} \ \frac{1}{\sqrt{s} + \frac{a}{2x^2}} \ du[/imath]

Well I still have [imath]\displaystyle x[/imath] inside the integral that I cannot rid of. Now I will take your advise and I will multiply both sides of the equation by [imath]\displaystyle x[/imath]. Let us see what happens.

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle ux = \sqrt{s}x^2 - \frac{a}{2}[/imath]

Would you say?

[imath]\displaystyle x^2 = \frac{ux + \frac{a}{2}}{\sqrt{s}} [/imath]

This will not help to get rid of the [imath]\displaystyle x[/imath]. (Please don't advise to solve that equation by quadratic formula. The integral will be so ugly and will get messy so that it is impossible to solve).

Therefore, what is your recommendation for the next step?

I didn't left the [imath]\displaystyle dx[/imath]. It is a standard way to write [imath]\displaystyle du = \left(\sqrt{s} + \frac{a}{2x^2}\right) \ dx \ \ [/imath] as [imath] \ \ \displaystyle du = \sqrt{s} + \frac{a}{2x^2} \ dx[/imath] in solving such problems. If you paid more attention, you wouldn't need to think the [imath]\displaystyle dx[/imath] was missing in the first term.
I recognize that many texts use
[imath]\displaystyle \int f(x) + g(x) \, dx[/imath]

when writing integrals. At no point has anyone written f(x) + g(x) dx without the integral. And if they did, they are very wrong. PEDMAS rules, even in Calculus.

Now, after seeing your confusion, I would disagree with these authors and my teacher and I am going to advise to write brackets whenever you have more terms.
Good. I don't recommend the notation, even inside the integral. It's just sloppy.

Now, I understand why you are angry and why you thought I don't know how do completing the square. And I also now understand why you have given me an example to show the method (thank you). You are saying if I know how to do that why I haven't shown it until the hint was given. First, I am used to do completing the square on equations that looks like [imath]\displaystyle ax^2 + bx[/imath] not [imath]\displaystyle ax + \frac{b}{x}[/imath], so I missed that it can be done on the latter as well. Also I was focusing a lot on choosing the correct substitution to get rid of the square root in the denominator. Therefore, I was distracted by these two things. Even experts like you may sometimes get distracted to recognize a simple idea in solving something.
You told me you didn't know how to solve [imath]u = \sqrt{s} x - \dfrac{a}{2 x}[/imath] for x. Just how much distraction are you under?

I think that you just scanned the problem and didn't actually solve it for [imath]\displaystyle u[/imath], this is why you recommend multiplying both sides by [imath]\displaystyle x[/imath]. I am sorry that I have skipped many steps which I thought that we reached to the level where we can see everything between the gaps. It seems that I have to show you all the steps, so that you can understand why multiplying both sides of the equation by [imath]\displaystyle x[/imath], will not work.
Yes. You need to show all of your steps. We've only been telling you this for ages. I'm glad you finally noticed the need.

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-\left(\sqrt{s}x - \frac{a}{2x}\right)^2} \ dx[/imath]

If we choose

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle du = \sqrt{s} + \frac{a}{2x^2} \ \ dx[/imath]

[imath]\displaystyle dx = \frac{1}{\sqrt{s} + \frac{a}{2x^2}} \ du[/imath]

[imath]\displaystyle \frac{2}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}e^{-u^2} \ \frac{1}{\sqrt{s} + \frac{a}{2x^2}} \ du[/imath]

Well I still have [imath]\displaystyle x[/imath] inside the integral that I cannot rid of. Now I will take your advise and I will multiply both sides of the equation by [imath]\displaystyle x[/imath]. Let us see what happens.

[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]

[imath]\displaystyle ux = \sqrt{s}x^2 - \frac{a}{2}[/imath]

Would you say?

[imath]\displaystyle x^2 = \frac{ux + \frac{a}{2}}{\sqrt{s}} [/imath]

This will not help to get rid of the [imath]\displaystyle x[/imath]. (Please don't advise to solve that equation by quadratic formula. The integral will be so ugly and will get messy so that it is impossible to solve).

Therefore, what is your recommendation for the next step?
My recommendation is not to do it this way at all. (I would do it by contour integration, myself.) There is, in fact, a way to manage this integral. (I am momentarily blanking on the method, but it's rather clever.) However, I would like to note something that has caused me to not get around to providing any sort of answer yet tonight. I had thought that you had moved from Bessel functions to Laplace transforms. I was slow to recognize it, but your integral is a modified Bessel function of the second kind.

[imath]K_{1/2}(a \sqrt{s} ) = \sqrt{\dfrac{\pi}{2}} \dfrac{e^{-a \sqrt{s}}}{\sqrt{a \sqrt{s}}}[/imath]

I'm not terribly familiar with these and I've been doing some research. As you probably got this problem from your text, I would recommend that you scour your materials for a method that you can use for this. Your source should have an integral representation of these that you can massage into the form given in the problem. I will get back to you when I get something more solid than what I have now.

-Dan

A, perhaps final, update from my end.

If we take the original integral:
[imath]\displaystyle \int_0^{\infty} \dfrac{e^{-st - a^2/(4t)}}{\sqrt{\pi t}} \, dt[/imath]

and look at the argument of the exponential
[imath]-s t - \dfrac{a^2}{4t} = -s \left ( t + \dfrac{a^2}{4s t} \right )[/imath]

[imath] = -\dfrac{a \sqrt{s}}{2} \left ( \left ( \dfrac{2 \sqrt{s}}{a} \right ) t + \dfrac{1}{ \left ( \dfrac{2 \sqrt{s}}{a} \right ) t} \right )[/imath]

and let
[imath]q = \left ( \dfrac{2 \sqrt{s}}{a} \right ) t[/imath]

which transforms the original integral into
[imath]\displaystyle \text{constant} \int_0^{\infty} \dfrac{e^{-(z/2)(q+1/q)}}{\sqrt{q}} \, dq[/imath]
where [imath]z = a \sqrt{s}[/imath]

Now we get a bit clever and substitute
[imath]t = e^y[/imath]

This gives
[imath]\displaystyle \text{constant} \int_0^{\infty} \dfrac{e^{-(z/2)(e^y+e^{-y})}}{e^{y/2}} \, e^y \,dy[/imath]

or
[imath]\displaystyle \text{constant} \int_0^{\infty} e^{y/2} e^{-z \, cosh(y)} \,dy[/imath]

Now, an integral involving [imath]e^{cosh(y)}[/imath] is the signature of a Bessel function, which is what got me thinking about them to do this integral.

As it happens
[imath]\displaystyle K_{1/2}(y) \propto \int_0^{\infty} cosh \left ( \dfrac{1}{2} y \right ) e^{-z \, cosh(y)} \, dy[/imath]

so we are almost there. By comparing the given solution with this Bessel integral, we can get a form for what the integral is supposed to be, but I can't find a way to finish it.

Or we could just look it up in a table and note that
[imath]\displaystyle K_{\nu}(z) = \dfrac{1}{2} \left ( \dfrac{z}{2} \right )^{\nu} \int_0^{\infty} \dfrac{e^{-( t + z^2/(4t) )}}{t^{\nu+1}} \, dt[/imath]

@mario99: I cannot find much on the net about modified Bessel functions of the second kind on the net. Could you please tell me what source you are working from? If this is a problem from a text, there is probably a clue given somewhere in the chapter on how to do this, however vague it might be. I might be able to intuit a better method by looking at your source.

-Dan

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My recommendation is not to do it this way at all. (I would do it by contour integration, myself.) There is, in fact, a way to manage this
I have done the contour integration before. The denominator of the integral always included a polynomial or a polynomial squared. I have never done it with a square root in the denominator. I tried to look for examples of this type, but I found nothing.

not get around to providing any sort of answer yet tonight. I had thought that you had moved from Bessel functions to Laplace transforms. I was slow to recognize it, but your integral is a modified Bessel function of the second kind.

[imath]K_{1/2}(a \sqrt{s} ) = \sqrt{\dfrac{\pi}{2}} \dfrac{e^{-a \sqrt{s}}}{\sqrt{a \sqrt{s}}}[/imath]
My Bessel skills are rising up these days and I will be so glad if this integral is really Bessel.

A, perhaps final, update from my end.

If we take the original integral:
[imath]\displaystyle \int_0^{\infty} \dfrac{e^{-st - a^2/(4t)}}{\sqrt{\pi t}} \, dt[/imath]

and look at the argument of the exponential
[imath]-s t - \dfrac{a^2}{4t} = -s \left ( t + \dfrac{a^2}{4s t} \right )[/imath]

[imath] = -\dfrac{a \sqrt{s}}{2} \left ( \left ( \dfrac{2 \sqrt{s}}{a} \right ) t + \dfrac{1}{ \left ( \dfrac{2 \sqrt{s}}{a} \right ) t} \right )[/imath]

and let
[imath]q = \left ( \dfrac{2 \sqrt{s}}{a} \right ) t[/imath]

which transforms the original integral into
[imath]\displaystyle \text{constant} \int_0^{\infty} \dfrac{e^{-(z/2)(q+1/q)}}{\sqrt{q}} \, dq[/imath]
where [imath]z = a \sqrt{s}[/imath]

Now we get a bit clever and substitute
[imath]t = e^y[/imath]

This gives
[imath]\displaystyle \text{constant} \int_0^{\infty} \dfrac{e^{-(z/2)(e^y+e^{-y})}}{e^{y/2}} \, e^y \,dy[/imath]

or
[imath]\displaystyle \text{constant} \int_0^{\infty} e^{y/2} e^{-z \, cosh(y)} \,dy[/imath]

Now, an integral involving [imath]e^{cosh(y)}[/imath] is the signature of a Bessel function, which is what got me thinking about them to do this integral.

As it happens
[imath]\displaystyle K_{1/2}(y) \propto \int_0^{\infty} cosh \left ( \dfrac{1}{2} y \right ) e^{-z \, cosh(y)} \, dy[/imath]

so we are almost there. By comparing the given solution with this Bessel integral, we can get a form for what the integral is supposed to be, but I can't find a way to finish it.

Or we could just look it up in a table and note that
[imath]\displaystyle K_{\nu}(z) = \dfrac{1}{2} \left ( \dfrac{z}{2} \right )^{\nu} \int_0^{\infty} \dfrac{e^{-( t + z^2/(4t) )}}{t^{\nu+1}} \, dt[/imath]

@mario99: I cannot find much on the net about modified Bessel functions of the second kind on the net. Could you please tell me what source you are working from? If this is a problem from a text, there is probably a clue given somewhere in the chapter on how to do this, however vague it might be. I might be able to intuit a better method by looking at your source.

-Dan
Thank you so much topsquark for the hard work. Manipulating the integral in that way was really smart. You have not got the exact form of the Bessel integral, but I still consider it winning because you have tried your best.

Let us assume that we were able to get the exact integral form of the modified Bessel function. When I do my calculations in the modified Bessel function, I get a different result.

[imath]\displaystyle[/imath]

[imath]\displaystyle K_n(x) = \frac{\pi}{2}\frac{I_{-n} - I_{n}}{\sin n\pi}[/imath]

[imath]\displaystyle K_{1/2}(a\sqrt{s}) = \frac{\pi}{2}\frac{I_{-1/2} - I_{1/2}}{\sin \pi/2}[/imath]

[imath]\displaystyle I_{1/2}(a\sqrt{s}) = \left(\frac{1}{2}a\sqrt{s}\right)^{1/2}\sum_{k=0}^{\infty}\frac{\left(\frac{1}{4}[a\sqrt{s}]^2\right)^k}{k!\Gamma(1/2 + k + 1)} = \left(\frac{1}{2}a\sqrt{s}\right)^{1/2}\frac{2\sinh(\sqrt{a}\sqrt[4]{s})}{\sqrt{\pi}\sqrt{a}\sqrt[4]{s}}[/imath]

[imath]\displaystyle I_{-1/2}(a\sqrt{s}) = \left(\frac{1}{2}a\sqrt{s}\right)^{-1/2}\sum_{k=0}^{\infty}\frac{\left(\frac{1}{4}[a\sqrt{s}]^2\right)^k}{k!\Gamma(-1/2 + k + 1)} = \left(\frac{1}{2}a\sqrt{s}\right)^{-1/2}\frac{\cosh(\sqrt{a}\sqrt[4]{s})}{\sqrt{\pi}}[/imath]

[imath]\displaystyle K_{1/2}(a\sqrt{s}) = \frac{\pi}{2}\left(\left[\frac{1}{2}a\sqrt{s}\right]^{-1/2}\frac{\cosh(\sqrt{a}\sqrt[4]{s})}{\sqrt{\pi}} - \left[\frac{1}{2}a\sqrt{s}\right]^{1/2}\frac{2\sinh(\sqrt{a}\sqrt[4]{s})}{\sqrt{\pi}\sqrt{a}\sqrt[4]{s}}\right) \neq \sqrt{\dfrac{\pi}{2}} \dfrac{e^{-a \sqrt{s}}}{\sqrt{a \sqrt{s}}}[/imath]

Even if we simplify the Hyperbolic functions to the exponential form, the result is not equal. I think that this is the reason why you could not get the wanted form of the integral.

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Even if we simplify the Hyperbolic functions to the exponential form, the result is not equal. I think that this is the reason why you could not get the wanted form of the integral.
I might agree with you...

except for that last line in my last post where I found your integral in a table...

If you won't give me your source, I'm done.

-Dan

I might agree with you...

except for that last line in my last post where I found your integral in a table...

If you won't give me your source, I'm done.

-Dan
This problem is not from a source. Most of the problems of the Laplace transform is done by referring to the table. I just like to prove what is in the table especially when the author says proving this identity is beyond the scope of the book.

This problem is not from a source. Most of the problems of the Laplace transform is done by referring to the table. I just like to prove what is in the table especially when the author says proving this identity is beyond the scope of the book.
Laplace transforms are taught in a Mathematical Methods class. The tables are simply a shortcut.

You are going to have to wait until
1. You find that new book
2. You have learned enough to understand it.

I'm only really speaking for myself, but as no one else has really chimed in much, I suspect that we can't help you with this any further.

At least for my part, I'm out.

-Dan

Addendum: This is not an official site policy statement but I would also like to point out that, whereas we are willing to help out with just about any Mathematics problem, you essentially posted a problem that you had no idea how to start and you didn't tell us anything about where it came from. We can usually help fix mistakes members make in their work, but we simply cannot teach members new material here: an online site is not a good venue to do this. If you have questions about completely new material to you, you would be better off hiring a face-to-face tutor or enrolling in an actual course at a college. Frankly, at the level of Mathematics that you are attempting to learn (whether or not you agree with me that you aren't ready for this level) a college course is really the only good way to learn this material.

You can, and probably will, ignore this statement as you usually ignore most of the advice that I give you.

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Laplace transforms are taught in a Mathematical Methods class. The tables are simply a shortcut.
I know.

Laplace transforms are taught in a Mathematical Methods class. The tables are simply a shortcut.

You are going to have to wait until
1. You find that new book
2. You have learned enough to understand it.
It seems that Mario is the one who is going to write this new book.

I'm only really speaking for myself, but as no one else has really chimed in much, I suspect that we can't help you with this any further.

At least for my part, I'm out.
You have already provided more than enough. I appreciate everything you have written and suggested.

Addendum: This is not an official site policy statement but I would also like to point out that, whereas we are willing to help out with just about any Mathematics problem, you essentially posted a problem that you had no idea how to start and you didn't tell us anything about where it came from.
This is not a new problem. I even have a solid idea about such problems. I have solved tons of problems involving this integral structure,

[imath]\displaystyle \int_{0}^{\infty}e^{-st}f(t) \ dt[/imath]

I have proved most of the Laplace table identities or whatever their names. You didn't ask me about the source of the problem when we were working in the first posts, and when you asked later, I answered.

It is true that this problem did not come from a source because no one ever asked to solve that integral, but the origin of the problem came from a Laplace transform table. Any differential calculus book, in the chapter of Laplace transform of the Error function, will include this identity in the table,

[imath]\displaystyle \mathscr{L}\bigg\{\frac{1}{\sqrt{\pi t}}e^{-\frac{a^2}{4t}}\bigg\} = \dfrac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

And by using the definition of Laplace transform, this problem becomes,

[imath]\displaystyle \mathscr{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t) \ dt = \dfrac{e^{-a\sqrt{s}}}{\sqrt{s}}[/imath]

where [imath]\displaystyle f(t) = \frac{1}{\sqrt{\pi t}}e^{-\frac{a^2}{4t}}[/imath]

which in turns is the integral in the first post.

You can, and probably will, ignore this statement as you usually ignore most of the advice that I give you.
Marios (2021, 2022, and 2023) have never ignored any statement, advice, suggestion, or anything else.

I have a feeling that this problem is not more difficult than the Airy equation. And because I have solved the Airy equation from scratch, I am capable of solving this integral. I am not saying it is going to be an easy task. It may even take hundreds of hours.

My current idea is that to expand the functions [imath]\displaystyle e^t[/imath] and [imath]\displaystyle \sqrt{t}[/imath] in infinite series. Then, inserting the correct form of them inside the integral and integrating term by term. Recognizing the solution of the infinite series and converting it to a nice function will be the most difficult part. Thanks to the Laplace transform table, it will make it the easiest part.

Thanks a lot Dan for the help and the hard work you have done in this post. You don't have to write anything after this post because I am going to end it at here.

This post should be locked out due to lack of interest.

Marios (2021, 2022, and 2023) have never ignored any statement, advice, suggestion, or anything else.
Ah. So you've stopped trying to learn new material and gone back to do a good review of Calculus I, like I suggested, so that you might actually be able to understand what you are doing instead of just regurgitating it.

I am so glad to hear that you are following that advice, as evidenced in this thread.

-Dan

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