How to compute $\int_{1}^{\infty} dx \sin \beta x e^{-\alpha x}x^n $or express it in terms of special functions?

Question

Compute $\int_{1}^{\infty} dx \sin \beta x e^{-\alpha x}x^n $, where $\beta ,\alpha >0$, $n$ is an integer (positive or negative).

Any help would be appreciated!

Answer 1

Deriving with respect to $\alpha$ $n$ times gets rid of that annoying $x^n$

$\int_1^\infty\sin(\beta x)e^{-\alpha x}x^n dx= (-1)^n\big(\frac{\partial}{\partial \alpha}\big)^n\int_1^\infty\sin(\beta x)e^{-\alpha x}dx$

Now lets define the following integrals

$I_{\sin} (\alpha,\beta)=\int_1^\infty\sin(\beta x)e^{-\alpha x}dx $

$I_{\cos} (\alpha,\beta)=\int_1^\infty\cos(\beta x)e^{-\alpha x}dx $

and define also this

$I(\alpha,\beta)=I_{\cos} + iI_{\sin}=\int_1^\infty [\cos(\beta x) +i\sin(\beta x)]e^{-\alpha x}dx=\int_1^\infty e^{-\alpha x + i\beta x }dx$

This integral is a simple exponential integrals and it converges for $\alpha>0$

$I (\alpha,\beta)=-\frac{e^{-\alpha + i\beta}}{-\alpha + i\beta}$

To get $I_{\sin}$ we just need to calculate the imaginary part of $I$

$I_{\sin} (\alpha,\beta) = \Im I (\alpha,\beta)= \frac{\alpha \sin\beta - \beta \cos \beta}{\alpha^2+\beta^2}e^{-\alpha}$

This means that you can write the original integral as

$\int_1^\infty\sin(\beta x)e^{-\alpha x}x^n dx=(-1)^{n}\big(\frac{\partial}{\partial \alpha}\big)^n\Big[\frac{\alpha \sin\beta - \beta \cos \beta}{\alpha^2+\beta^2}e^{-\alpha}\Big]$

And I think you can stop here because deriving is just a mechanical task, and the result will be ugly (lots of sums coefficients and binomial coefficients)

Answer 2

The integral can be rephrased to $$I_n(z)= Im \int_1^{+\infty}e^{zx} x^n dx$$ where $z= -\alpha + i\beta$ with $\alpha,\beta >0$. Now observe that $$I_0(z)= Im \int_1^{+\infty}e^{zx} dx= -Im \frac{e^z}{z} \:.$$ On the other hand $$\frac{d^n}{dz^n}I_0(z)= Im \int_1^{+\infty}x^n e^{zx} dx = I_n(z)\:.$$ In summary, $$\int_1^{+\infty}\sin(\beta x) e^{-\alpha x} x^n dx= Im \left.\frac{d^n}{dz^n}\right|_{z= -\alpha + i \beta} \frac{e^z}{z} = Im\: e^z \left.\sum_{k=0}^n (-1)^k {{n}\choose{k}} k! z^{-(k+1)}\right|_{z= -\alpha + i \beta}$$