A contest math integral: $\int_1^\infty \frac{\text{d}x}{\pi^{nx}-1}$ My school holds a math contest that has problems that vary level to level. Nobody managed to solve this particular one:
$$\int_1^\infty \frac{\text{d}x}{\pi^{nx}-1}$$
In terms of $n$
I was wondering if there is a solution to this integral?
 A: Put $u = \pi^{nx}-1$
You will get to point where the integral will be
$$I = \frac{1}{n\log\pi}\int \left[\frac{1}{u} - \frac{1}{1+u}\right]du$$
$$I = \frac{1}{n\log\pi} \ln\left(\frac{u}{1+u}\right)$$
Evaluate and you get the result.
A: Mathematica told me there is no closed form, but I found this working out the integral:
Assume that $m$ and $n$ are positive $\to m,n\in\mathbb{R^+}$:
$$\text{I}=\int_{1}^{\infty}\frac{1}{\pi^{nx}-1}\space\text{d}x=$$
$$\lim_{m\to\infty}\int_{1}^{m}\frac{1}{\pi^{nx}-1}\space\text{d}x=$$

Substitute $u=nx$ and $\text{d}u=n\space\text{d}x$:
This gives a new lower bound $u=n\cdot1=n$ and upper bound $u=n\cdot m=mn$:

$$\lim_{m\to\infty}\frac{1}{n}\int_{n}^{mn}\frac{1}{\pi^{u}-1}\space\text{d}u=$$

Substitute $s=\pi^u$ and $\text{d}s=\pi^u\ln(\pi)\space\text{d}u$:
This gives a new lower bound $s=\pi^{n}$ and upper bound $s=\pi^{mn}$:

$$\lim_{m\to\infty}\frac{1}{n\ln(\pi)}\int_{\pi^{n}}^{\pi^{mn}}\frac{1}{s(s-1)}\space\text{d}s=$$
$$\lim_{m\to\infty}\frac{1}{n\ln(\pi)}\int_{\pi^{n}}^{\pi^{mn}}\left(\frac{1}{s-1}-\frac{1}{s}\right)\space\text{d}s=$$
$$\lim_{m\to\infty}\frac{1}{n\ln(\pi)}\left(\int_{\pi^{n}}^{\pi^{mn}}\frac{1}{s-1}\space\text{d}s-\int_{\pi^{n}}^{\pi^{mn}}\frac{1}{s}\space\text{d}s\right)=$$
$$\lim_{m\to\infty}\frac{1}{n\ln(\pi)}\left(\int_{\pi^{n}}^{\pi^{mn}}\frac{1}{s-1}\space\text{d}s-\left[\ln\left|s\right|\right]_{\pi^{n}}^{\pi^{mn}}\right)=$$

Substitute $p=s-1$ and $\text{d}p=\text{d}s$:
This gives a new lower bound $p=\pi^{n}-1$ and upper bound $p=\pi^{mn}-1$:

$$\lim_{m\to\infty}\frac{1}{n\ln(\pi)}\left(\int_{\pi^{n}-1}^{\pi^{mn}-1}\frac{1}{p}\space\text{d}p-\left[\ln\left|s\right|\right]_{\pi^{n}}^{\pi^{mn}}\right)=$$
$$\lim_{m\to\infty}\frac{1}{n\ln(\pi)}\left(\left[\ln\left|p\right|\right]_{\pi^{n}-1}^{\pi^{mn}-1}-\left[\ln\left|s\right|\right]_{\pi^{n}}^{\pi^{mn}}\right)=$$
$$\lim_{m\to\infty}\frac{\ln\left|\pi^{mn}-1\right|-\ln\left|\pi^{n}-1\right|-\ln\left|\pi^{mn}\right|+\ln\left|\pi^{n}\right|}{n\ln(\pi)}=$$
$$\frac{1}{n\ln(\pi)}\lim_{m\to\infty}\left(\ln\left|\pi^{mn}-1\right|-\ln\left|\pi^{n}-1\right|-\ln\left|\pi^{mn}\right|+\ln\left|\pi^{n}\right|\right)=$$
$$\frac{1}{n\ln(\pi)}\lim_{m\to\infty}\ln\left(\frac{\pi^n-\pi^{n-mn}}{\pi^n-1}\right)=\frac{\ln\left(1+\frac{1}{\pi^n-1}\right)}{n\ln(\pi)}$$
A: Observe this very simple step: $\int_1^\infty \frac{\text{d}x}{\pi^{nx}-1}$=$\int_1^\infty \frac{1-\pi^{nx}+\pi^{nx}}{\pi^{nx}-1}dx$=$-\int_1^\infty1dx$+$\int_1^\infty \frac{\pi^{nx}}{\pi^{nx}-1}dx$. Derivative of denominator is in the numerator
