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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?

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  • $\begingroup$ @Sky It converges for $n\ge 0$ $\endgroup$
    – user266519
    Dec 23, 2015 at 19:27
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    $\begingroup$ The $\pi$ is a disguise. You have $1 / (e^{Tx} - 1)$ for real constant $T > 0$ $\endgroup$
    – Will Jagy
    Dec 23, 2015 at 19:35
  • $\begingroup$ @Apple Yes now it converges with the changed lower limit of Integration $\endgroup$
    – meiji163
    Dec 23, 2015 at 19:47

3 Answers 3

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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.

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  • $\begingroup$ Ohhh Why didn't I think of that? $\endgroup$
    – user266519
    Dec 23, 2015 at 19:29
  • $\begingroup$ That is OK. As you start to solve many problems, you will be at ease on substitution $\endgroup$ Dec 23, 2015 at 19:30
  • $\begingroup$ I tried relating this to the zeta function $\endgroup$
    – user266519
    Dec 23, 2015 at 19:40
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    $\begingroup$ As I commented, the $\pi$ is a disguise,, it is really $$ \frac{1}{e^{Tx} - 1}$$ for constant real $T > 0.$ $\endgroup$
    – Will Jagy
    Dec 23, 2015 at 19:43
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    $\begingroup$ @WillJagy, you are absolutely right!!. $\endgroup$ Dec 23, 2015 at 19:44
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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 $p=\pi^{mn}-1$:


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

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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

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