About the integrals $\int_{0}^{+\infty}\sin(x)^k e^{-x}dx$

Question

The integral: $$J(x,k)=\int_{0}^{x}\dfrac{\sin(z)^k}{\exp(z)}dz$$ is: $$J(x,k)=\frac{e^{-x} \left(2-2 e^{2 i x}\right)^{-k} \left(-i e^{-i x} \left(-1+e^{2 i x}\right)\right)^k \, _2F_1\left(\frac{i}{2}-\frac{k}{2},-k;\left(1+\frac{i}{2}\right)-\frac{k}{2};e^{2 i x}\right)}{-1-i k}$$ I am stuck to find $$\lim_{x\to+\infty}J(x,k)=\int_{0}^{+\infty}\sin(z)^k e^{-z}\,dz.$$ Using Mathematica I get $J(\infty,1),J(\infty,2),J(\infty,3)$ and so on, but I would know if is it possible to find a general expression for $k\in\mathbb{N}$. Thanks in advance.

Answer 1

Since $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\int_{0}^{+\infty}e^{(ri-1)x}\,dx = \frac{1}{1-ri}$ we have:

$$ I_n=\int_{0}^{+\infty}\sin(x)^n e^{-x}\,dx = \frac{1}{(2i)^n}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{1+i(2k-n)} \tag{1}$$ and by Herglotz' trick

$$ I_n = \frac{\pi n!}{2^{n+1}\Gamma\left(\frac{n+2-i}{2}\right)\Gamma\left(\frac{n+2+i}{2}\right)}\cdot\left\{\begin{array}{rcl}\text{csch}\frac{\pi}{2}&\text{if}&n\text{ is even}\\\text{sech}\frac{\pi}{2}&\text{if}&n \text{ is odd}\end{array}\right.\tag{2} $$ or, in a more compact form, $$ I_n = n!\prod_{k=0}^{\lfloor n/2\rfloor}\left(1+(n-2k)^2\right)^{-1}\tag{3}$$ that can be achieved by repeated integration by parts, too. Thanks to user90369 for the suggestion.