calculate $\int_0^\infty x^n\sin(2\pi \ln(x))e^{-\ln^2(x)} \mathrm{d}x$ I have to show, that $$\int_0^\infty x^n\sin(2\pi \ln(x))e^{-\ln^2(x)} \mathrm{d}x=0$$ and I know, that $\int_{-\infty}^{\infty} \sin(2\pi t)e^{-t^2} \mathrm{d}t=0$.
So, I thought of substitution:
$t=\ln(x) \Leftrightarrow x=e^t$, so $x \mathrm{d}t=\mathrm{d}x$, so we get:
$$\int_{-\infty}^{\infty} e^{t(n+1)}\sin(2\pi t)e^{-t^2} \mathrm{d}t$$
Now I did integration by parts with $u=\sin(2\pi t)e^{-t^2}$ and $v'=e^{t(n+1)}$,
but I don't see that this is $0$. Maybe someone can help me.
 A: Hint: $\displaystyle t^2-t(n+1)=(t-\frac{n+1}{2})^2-\frac{(n+1)^2}{4}$. So try to put $\displaystyle u=t-\frac{n+1}{2}$ 
A: Begin by enforcing the substitution $x\to e^x$.  Then, we have
$$\begin{align}
\int_0^\infty x^n\sin(2\pi \log(x))e^{-\log^2(x)}\,dx&=\int_{-\infty}^\infty e^{-x^2+(n+1)x}\sin (2\pi x)\,dx\\\\
&=e^{(n+1)^2/4}\int_{-\infty}^{\infty}e^{-\left(x-\frac{n+1}{2}\right)^2}\sin (2\pi x)\,dx\\\\
&=(-1)^{n+1}e^{(n+1)^2/4}\int_{-\infty}^{\infty}e^{-x^2}\sin (2\pi x)\,dx\\\\
&=0
\end{align}$$
where we exploited the fact that the integrand is an odd function.  
A: So then: $u=t-\frac{n+1}{2}$ and $du=dt$.
So we get $\int_{-\infty}^{\infty} e^{-u^2+\frac{(n+1)^2}{4}} \sin(2\pi(u+\frac{n+1}{2})) \mathrm{d}u$. Because of the angle addition theorems, we can write:
$e^{\frac{(n+1)^2}{4}} \cos(2\pi \frac{n+1}{2}) \int_{-\infty}^{\infty} e^{-u^2} \sin(2\pi u) du + e^{\frac{(n+1)^2}{4}} \sin(2\pi \frac{n+1}{2}) \int_{-\infty}^{\infty} e^{-u^2} \cos(2\pi u) du$
So we know, that $\int_{-\infty}^{\infty} e^{-u^2} \sin(2\pi u) du =0$, but $\int_{-\infty}^{\infty} e^{-u^2} \cos(2\pi u) du \ne0$.
Sorry, but what is wrong here?
A: 
Now I did integration by parts with $u=\sin(2\pi t)e^{-t^2}$ and $v'=e^{t(n+1)}$

Why ? :-\ Just use Euler's formula in conjunction with the value of the Gaussian integral. All you have to do is to complete the square.
