Is there any solution except integration by parts for $\int_0^\infty x^2 e^{-x} \sin(\alpha x) dx$? I want to calculate the Fourier sine integral and the Fourier cosine integral of the function $f(x)=x^2e^{-x}$. So I have to calculate:
$$
\int_0^\infty x^2e^{-x}\sin(\alpha x)dx \qquad , \qquad \int_0^\infty x^2e^{-x}\cos(\alpha x)dx
$$
The integration by parts method is really complex for these two integrals! 
Is there any better solution to calculate them?!
 A: $$\int_{0}^{+\infty}x^2 e^{-(1-i\alpha)x}\,dx=\frac{1}{(1-i\alpha)^3}\int_{0}^{+\infty}x^2 e^{-x}\,dx =\frac{2}{(1-i\alpha)^3}=\frac{2}{(1+\alpha^2)^3}(1+i\alpha)^3 $$
then you may just consider the real or imaginary parts of both sides to get:
$$ \int_{0}^{+\infty}x^2 e^{-x}\cos(\alpha x)\,dx = \frac{2-6\alpha^2}{(1+\alpha^2)^3},\qquad \int_{0}^{+\infty}x^2 e^{-x}\sin(\alpha x)\,dx = \frac{6\alpha-2\alpha^3}{(1+\alpha^2)^3}.$$
A: One might consider the following integral:
$$f(\alpha)=\int_0^\infty e^{-x}\sin(\alpha x)~\mathrm dx$$
This can be solved using Euler's formula:
$$\begin{align}f(\alpha)&=\Im\int_0^\infty e^{(i\alpha-1)x}~\mathrm dx\\&=\Im\left[\frac{e^{(i\alpha-1)x}}{i\alpha-1}\right]_0^\infty\\&=\Im\left[\frac1{1-i\alpha}\right]\end{align}$$
Likewise, from the integral representation:
$$f''(\alpha)=-\int_0^\infty x^2e^{-x}\sin(\alpha x)~\mathrm dx$$
And from the closed form:
$$f''(\alpha)=\Im\left[\frac{-2}{(i\alpha-1)^3}\right]=\Im\left[\frac{2(i\alpha+1)^3}{(\alpha^2+1)^3}\right]=\frac{2\alpha^3-6\alpha}{(\alpha^2+1)^3}$$
Likewise,
$$\begin{align}g(\alpha)&=\int_0^\infty e^{-x}\cos(\alpha x)~\mathrm dx\\&=\Re\int_0^\infty e^{(i\alpha-1)x}~\mathrm dx\\&=\Re\left[\frac1{1-i\alpha}\right]\\\implies f''(\alpha)&=\Re\left[\frac{2(i\alpha+1)^3}{(\alpha^2+1)^3}\right]=\frac{2-6\alpha^2}{(\alpha^2+1)^3}\end{align}$$
$$f''(\alpha)=-\int_0^\infty x^2e^{-x}\cos(\alpha x)~\mathrm dx$$
A: Laplace transform 
$\mathcal{L}\{x^2\sin\alpha x\}(s)=-\dfrac{2 a \left(a^2-3 s^2\right)}{\left(a^2+s^2\right)^3}$
your integral is
$\mathcal{L}\{x^2\sin\alpha x\}(1)=-\dfrac{2 a \left(a^2-3\right)}{\left(a^2+1\right)^3}$
For cosine is
$\mathcal{L}\{x^2\cos\alpha x\}(1)=\dfrac{2 \left(1-3 a^2\right)}{\left(a^2+1\right)^3}$
A: If you don't want to use complex numbers, you'll find $\int_0^\infty e^{-\beta x}\cos\alpha x$ and its sine counterpart are relatively easy by parts. Then just apply $\partial_\beta^2$ and set $\beta=1$.
