Compute $\int_{0}^{e}\sin(\pi\ln(x))dx$ I've got this improper integral (calculating the value) and I have difficulties understanding the integration steps
$$\int_{0}^{e}\sin(\pi\ln(x))dx$$
I know that $\sin(\pi)=0$ and $\ln(0)$ is not defined.
So, I would use substitution, with $u=\ln(x)$ and $du=\frac{1}{x}dx$ and adjust the bounds:
$$\int_{\ln(0)}^{\ln(e)} e^u\sin(\pi u) du = \int_{-\infty}^1 e^u\sin(\pi u) du$$  
My question is, how do I get the $e^u$? 
And then, how do I continue? 
For continuation I would use partial integration to calculate the integral, or this there a kind of trick/rule to calculate it?
 A: If $u=\log x$, so that $du=\frac{1}{x}\,dx$, then $dx = x\,du = e^u\,du$. 
To continue, integration by parts gives:
$$ I=\int_{-\infty}^{1}e^u\,sin(\pi u)\,du = -\pi\int_{-\infty}^{1}e^u\cos(\pi u)\,du=\pi e-\pi^2\int_{-\infty}^{1}e^u\sin(\pi u)\,du\tag{1}$$
and $I=\pi e-\pi^2 I $ leads to:
$$ I = \color{red}{\frac{\pi e}{1+\pi^2}}.\tag{2}$$
A: We have the integral we want to evaluate:
$$I:=\int_{0}^{e}\sin(\pi \ln(x))\:\mathrm{d}x$$
We can make the substitution $u = \ln(x)$ and therefore $\mathrm{d}u = \frac{\mathrm{d}x}{x} \implies \mathrm{d}x = e^{u}\:\mathrm{d}u$. This gives us:
$$I=\int_{-\infty}^{1}e^{u}\sin(\pi u)\:\mathrm{d}u$$
We can now integrate by parts with $w = \sin(\pi u)$, $v' = e^{u}$:
$$I=\left.e^{u}\sin(\pi u)\right|_{u=-\infty}^{1}-\pi\int_{-\infty}^{1}e^{u}\cos(\pi u)\:\mathrm{d}u$$
Integrating the integral by parts again, we get:
$$I = \left.e^{u}\sin(\pi u)\right|_{u=-\infty}^{1}-\left.\pi e^{u}\cos(\pi u)\right|_{u=-\infty}^{1} - \pi^{2}\int_{-\infty}^{1}e^{u}\sin(\pi u)\:\mathrm{d}u$$
But the last integral is just $I$ again, so we have:
$$(1+\pi^{2})I = \left.e^{u}\sin(\pi u)\right|_{u=-\infty}^{1}-\left.\pi e^{u}\cos(\pi u)\right|_{u=-\infty}^{1} = \pi e$$
Where we note that as $u\to-\infty$ the $e^{u}$ dominates, bringing the function to $0$. Therefore:
$$I=\frac{\pi e}{1+\pi^{2}}$$
