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

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 1 '15 at 17:26
• @mickep. I revised it. Also can you provide a basic book about how to integrate above integral? – Jack Dec 2 '15 at 2:25

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

• Please see the changed – Jack Dec 2 '15 at 2:28
• @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 2 '15 at 2:45
• What do you mean by a simple linear substitution? Change variable($y=x(1/\lambda-it)$)? – Jack Dec 2 '15 at 2:58
• @Jack: Yes, exactly. – Lucian Dec 2 '15 at 3:05
• 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 2 '15 at 3:08

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

• I have not been told how to integrate complex number. – Jack Dec 1 '15 at 15:59

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