How to integrate $\int_{0}^{\infty}{\cos(tx)x^{\alpha-1}e^{-x/\lambda}}dx$ and $\int_{0}^{\infty}{\sin(tx)x^{\alpha-1}e^{-x/\lambda}}dx$

Question

The original problem is $\int_{0}^{\infty}{e^{itx}x^{\alpha-1}e^{-x/\lambda}}dx$

My work:

$\int_{0}^{\infty}{\cos(tx)x^{\alpha-1}e^{-x/\lambda}}dx=-\lambda\int_{0}^{\infty}{\cos(tx)x^{\alpha-1}de^{-x/\lambda}}=-\lambda t\int_{0}^{\infty} {\sin(tx)x^{\alpha-1}e^{-x/\lambda}}dx+\lambda (\alpha-1)^2\int_{0}^{\infty}{\cos(tx)x^{\alpha-2}e^{-x/\lambda}}dx.$

Changed Can I solve it this way? Let $y=x(1/\lambda-it)$. Solve this $\int_{0}^{\infty}{e^{itx}x^{\alpha-1}e^{-x/\lambda}}dx$. My concern is when $x=\infty$, is $y$ also infinity?

I do not how to continue.

From your comment on Chinny84's answer, I must ask: What tools do you have? It would be easier and it could save time and effort if you told so in the question. Also, what is the source of this integral? Is it an exercise from a book? — mickep, Dec 01 '15 at 17:26
@mickep. I revised it. Also can you provide a basic book about how to integrate above integral? — Jack, Dec 02 '15 at 02:25

score 2 · Answer 1 · answered Dec 01 '15 at 17:22

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Hint: With the help of a simple linear substitution, try to express or rewrite your integral in terms of the famous $\Gamma$ function.

answered Dec 01 '15 at 17:22

Lucian

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Please see the changed – Jack Dec 02 '15 at 02:28
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@Jack: Technically, it will be an oriented, complex infinity. But it can be proven, using complex integration along a certain contour, that $\displaystyle\int_0^\infty=\int_0^{z~\infty}.~$ However, doing that defeats the purpose of simplicity. $($Personally, I prefer a symbolic, intuitive approach to a very rigorous one$).$ – Lucian Dec 02 '15 at 02:45
What do you mean by a simple linear substitution? Change variable($y=x(1/\lambda-it)$)? – Jack Dec 02 '15 at 02:58
@Jack: Yes, exactly. – Lucian Dec 02 '15 at 03:05
So This is the problem I have. After changing variable, we need the change the upper limit and lower limit of integral. Lower limit is $0$, but how do you define $y$ when $x\rightarrow\infty$? And I do not understand $\int^{z\infty}_{0}$? What is $z\infty$ – Jack Dec 02 '15 at 03:08
@Jack: I already answered. $~z=\dfrac1\lambda-it,~$ and $z~\infty$ is the oriented infinity along the line uniting the origin with z. – Lucian Dec 02 '15 at 03:13
Can you provide more details about why $\int_{0}^{\infty}=\int_{0}^{z\infty}$? – Jack Dec 02 '15 at 22:13
@Jack: Use techniques similar to the ones presented in these three posts. – Lucian Dec 02 '15 at 23:08

score 1 · Answer 2 · answered Dec 01 '15 at 15:52

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$$ \int \cos(tx)x^{\alpha-1}\mathrm{e}^{-x/\lambda}dx= \mathcal{Re}\left(\int \mathrm{e}^{itx}x^{\alpha-1}\mathrm{e}^{-x/\lambda}\right) = \mathcal{Re}\left(i^{\alpha-1}\dfrac{d^{\alpha-1}}{dt^{\alpha-1}}\int \mathrm{e}^{itx}\mathrm{e}^{-x/\lambda}dx\right) $$ This could be an easier approach.

answered Dec 01 '15 at 15:52

Chinny84

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I have not been told how to integrate complex number. – Jack Dec 01 '15 at 15:59

score 0 · Answer 3 · edited Dec 01 '15 at 16:34

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Rewrite the cosine as a sum of two complex exponentials, then simply integrate $\alpha$ times by parts.

edited Dec 01 '15 at 16:34

Alex M.

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answered Dec 01 '15 at 15:47

Nikita Astrakhantsev

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How to integrate $\int_{0}^{\infty}{\cos(tx)x^{\alpha-1}e^{-x/\lambda}}dx$ and $\int_{0}^{\infty}{\sin(tx)x^{\alpha-1}e^{-x/\lambda}}dx$

3 Answers3