Determine the values of the real number [MATH]\alpha[/MATH] for which the integral converges

[Like Tony Stark]

[MATH] \int_{0}^{\infty}{\frac{cosx}{x^2+\alpha^2} dx} [/MATH]
The integral is improper because it tends to infinity, there are no points of discontinuity since the domain of the function is R.
I had the idea of solving the integral so that I can notice the values that would make it convergent when it tends to infinity, but the thing is that I don't know how to solve it.

 

[pka]

Look at this link. That is a nasty integral for \(\alpha=2\).
You need to tell us what topics you have studied in numerical methods.
Give us some idea what you know about this in general.

 

[lookagain]

[MATH] \int_{0}^{\infty}{\frac{cosx}{x^2+\alpha^2} dx} [/MATH]
The integral is improper because it tends to infinity, there are no points of discontinuity since the domain of the function is R.

If alpha were to equal 0, then the denominator would be \(\displaystyle \ x^2 \). You cannot
have x be 0, so that the domain of the function would not be all real numbers in
that case. Are you told that alpha is real and positive?

 

[Like Tony Stark]

Look at this link. That is a nasty integral for \(\alpha=2\).
You need to tell us what topics you have studied in numerical methods.
Give us some idea what you know about this in general.

We're studying improper integrals, integrals with parameters, gamma function, etc. No complex analysis

 

[Like Tony Stark]

If alpha were to equal 0, then the denominator would be \(\displaystyle \ x^2 \). You cannot
have x be 0, so that the domain of the function would not be all real numbers in
that case. Are you told that alpha is real and positive?

It just says that α can be any real number

 

[pka]

If alpha were to equal 0, then the denominator would be \(\displaystyle \ x^2 \). You cannot
have x be 0, so that the domain of the function would not be all real numbers in that case. Are you told that alpha is real and positive?

Is that true?
Surely you understand that \(\displaystyle\int_0^1 {\frac{1}{{\sqrt x }}dx}=2 ~?\) SEE HERE

 

[lookagain]

Is that true?
Surely you understand that \(\displaystyle\int_0^1 {\frac{1}{{\sqrt x }}dx}=2 ~?\) SEE HERE

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I was stating that the domain of the function \(\displaystyle \ \dfrac{cos(x)}{x^2} \ \) is not all real numbers. It is not continuous at x = 0. The OP claimed there are no points of discontinuity. If alpha were to equal 0, then the function would be
discontinuous. Being discontinuous might not keep it from having a finite sum, but I was not addressing that. (edits)

 

[pka]

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I was stating that the domain of the function \(\displaystyle \ \dfrac{cos(x)}{x^2} \ \) is not all real numbers. It is not continuous at x = 0. The OP claimed there are no points of discontinuity. If alpha were to equal 0, then the function would be
discontinuous. That might not keep it from having a finite sum, but I was
addressing that.

Well what is the domain of \(\dfrac{1}{\sqrt x}~?\)

 

[lookagain]

Well what is the domain of \(\dfrac{1}{\sqrt x}~?\)

It's x > 0, and that integral you involved it with happens to converge.