Did you try substitution:[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.[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?
When [imath]\displaystyle x = \sqrt{t}[/imath], the integral becomesDid you try substitution:
\(\displaystyle \sqrt{t} = x\)
I have this ideaI 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
Are you saying that you don't know how to integrateAccording 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].
C'mon! Partial differential equations and you can't take a simple diffferential?[imath]\displaystyle u = \sqrt{s}x - \frac{a}{2x}[/imath]
[imath]\displaystyle du = \sqrt{s} \, \red{\textbf{dx}}+ \frac{a}{2x^2} \ \ dx[/imath]
Come on Dan. I didn't ask about this part. I have already done it.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
Come on Dan. This is a simple derivative.C'mon! Partial differential equations and you can't take a simple diffferential?
(No, this doesn't help you solve the problem, but please check your work!)
-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.Come on Dan. I didn't ask about this part. I have already done it.
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 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 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.
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.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.
Yes.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.
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.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 recognize that many texts useI 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.
Good. I don't recommend the notation, even inside the integral. It's just sloppy.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.
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?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.
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.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.
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]\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 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.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
My Bessel skills are rising up these days and I will be so glad if this integral is really Bessel.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]
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.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
I might agree with you...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.
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.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
Laplace transforms are taught in a Mathematical Methods class. The tables are simply a shortcut.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.
I know.Laplace transforms are taught in a Mathematical Methods class. The tables are simply a shortcut.
It seems that Mario is the one who is going to write this new 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.
You have already provided more than enough. I appreciate everything you have written and suggested.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.
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,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.
Marios (2021, 2022, and 2023) have never ignored any statement, advice, suggestion, or anything else.You can, and probably will, ignore this statement as you usually ignore most of the advice that I give you.
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.Marios (2021, 2022, and 2023) have never ignored any statement, advice, suggestion, or anything else.
This post should be locked out due to lack of interest.