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$$\int_0^\infty \frac{x^n\mathrm dx}{e^{nx}-1}= \frac{\Gamma(n)}{n^n}\zeta{(n+1)}$$

I have evaluated the integral by turning denominator into a geometric series and by switching integral and summation. However, my actual answer was in terms of factorials and I had assumed that $n\in\mathbb N$ while solving to make it into this form. Using an online calculator, I had verified it for $n=1,2,3$.

Questions: Can we solve it by any other method ? For what (if any) does it holds for some non-natural $n$ ?

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  • $\begingroup$ This approach is correct. $\Gamma(n)$ will always make sense if $n>0$, this identity is always correct if $n>0$. You should find that your working will, well, work, if $n$ is not integer $\endgroup$
    – FShrike
    Commented Jun 26, 2023 at 11:13
  • $\begingroup$ @FShrike Thanks. But as said in question, I actually arrived at answer in factorial form i.e. $\frac{\zeta{(n+1)(n-1)!}}{n^n}$for simplicity of evaluation and only switched to gamma later for a neat look. $\endgroup$ Commented Jun 26, 2023 at 11:18
  • $\begingroup$ Why wouldn't your approach work if $n$ is not an integer ? $\endgroup$ Commented Jun 26, 2023 at 11:24
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    $\begingroup$ @SineoftheTime Yes I get it. $\endgroup$ Commented Jun 26, 2023 at 17:50
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    $\begingroup$ @An_Elephant By taking $t=xn$, your integral becomes $$ \frac{1}{{n^{n + 1} }}\int_0^{ + \infty } {\frac{{t^n }}{{{\rm e}^t - 1}}{\rm d}t} . $$ Now use the result in your closed question. $\endgroup$
    – Gary
    Commented Jun 27, 2023 at 5:11

3 Answers 3

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We can use a slight extension of Ramanujan's master theorem. I will include a proof below.

Assume $f$ and $h$ have an expansion of the form, $$f(x)=\sum_{m=1}^\infty h(mx),\quad h(x)=\sum_{k=0}^\infty\frac{\varphi(k)}{k!}(-x)^k,$$ then the Mellin transform of $f(x)$ is given by, $$\int_0^{+\infty} x^{n-1}f(x)\ dx=\Gamma(n)\zeta(n)\varphi(-n).$$

In our case, $$f(x)=\frac{1}{e^{nx}-1}=\frac{1}{e^{nx}}\frac{1}{1-e^{-nx}}=\frac{1}{e^{nx}}\sum_{k=0}^\infty e^{-knx}=\sum_{k=1}^\infty e^{-knx}$$ where, $$h(kx)=e^{-knx}=\sum_{m=0}^\infty \frac{n^m}{m!}(-kx)^m$$ hence the Mellin transform, $$\int_0^{+\infty} \frac{x^{s-1}}{e^{nx}-1}\ dx=\Gamma(s)\zeta(s)n^{-s}$$ take $s=n+1$, $$\int_0^{+\infty} \frac{x^n}{e^{nx}-1}\ dx=\frac{\Gamma(n+1)}{n^{n+1}}\zeta(n+1)=\frac{\Gamma(n)}{n^n}\zeta(n+1).$$

Proof. By Ramanujan's master theorem, $$\int_0^{+\infty}x^{n-1} h(x)\ dx=\Gamma(n)\varphi(-n)$$ subbing $x\mapsto mx$, $$m^{n}\int_0^{+\infty} x^{n-1}h(mx)\ dx=\Gamma(n)\varphi(-n)$$ rearranging and summing over $m\in\mathbb{N}$, $$\int_0^{+\infty} x^{n-1}\sum_{m=1}^\infty h(mx)\ dx=\sum_{m=1}^\infty m^{-n}\Gamma(n)\varphi(-n)=\Gamma(n)\zeta(n)\varphi(-n)$$ hence, $$\int_0^{+\infty} x^{n-1}f(x)\ dx=\Gamma(n)\zeta(n)\varphi(-n).$$

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    $\begingroup$ Awesome!! I will need to tuck this away into my integral solving toolbox. $\endgroup$
    – K.defaoite
    Commented Aug 14, 2023 at 14:27
  • $\begingroup$ @K.defaoite. If you like this, you might also like my question on generalizing the master theorem. (math.stackexchange.com/q/4744295/1141581) $\endgroup$
    – bob
    Commented Aug 14, 2023 at 14:53
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Alternatively, we can use generating functions and bracket integration.

By the generating function for Bernoulli numbers $B_k$, $$\frac{x}{e^x-1}=\sum_{k=0}^\infty\frac{B_k}{k!}x^k$$ we have, $$\frac{x^n}{e^{nx}-1}=\sum_{k=0}^\infty\frac{B_kn^k}{k!n}x^{k+n-1}=\sum_{k=0}^\infty\frac{(-1)^{k+1}n^kk\zeta(1-k)}{k!n}x^{k+n-1}=-\sum_{k=0}^\infty\phi_kn^{k-1}k\zeta(1-k)x^{k+n-1}$$ here I used relation (43) and the indicator $\phi_k$, $$B_k=(-1)^{k+1}k\zeta(1-k),\quad \phi_k=\frac{(-1)^k}{k!},$$ re-expressing the integral as a bracket series, $$\int_0^{+\infty}\frac{x^n}{e^{nx}-1}\ dx=-\sum_{k=0}^\infty\phi_kn^{k-1}k\zeta(1-k)\langle k+n\rangle$$ which assigns the value, $$-n^{-n-1}{(-n)}\zeta(1+n)\Gamma(n)=\frac{\Gamma(n)}{n^n}\zeta(1+n).$$

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Let $t=nx\implies dt=ndx$, we have: $$\int_0^\infty \frac{x^n\mathrm dx}{e^{nx}-1}=\int_{0}^{\infty}\left(\frac{t}{n}\right)^n\cdot\frac{1}{e^t-1}\cdot \frac1n dt=\frac{1}{n^{n+1}}\cdot\int_{0}^{\infty}\frac{t^n}{e^t-1}dt$$ We are going to use power series expansion of the denominator, so: $$\frac{1}{n^{n+1}}\cdot\int_{0}^{\infty}\frac{t^n}{e^t-1}dt=\frac{1}{n^{n+1}}\cdot\int_{0}^{\infty}\frac{t^n\cdot e^{-t}}{1-e^{-t}}dt$$ In particular: $$\frac{1}{1-e^{-t}}=\sum_{i=0}^{\infty}e^{-it}$$ So: $$\frac{1}{n^{n+1}}\cdot\int_{0}^{\infty}t^n\cdot e^{-t}\sum_{i=0}^{\infty}e^{-it}dt=\frac{1}{n^{n-1}}\cdot\sum_{i=0}^{\infty}\int_{0}^{\infty}t^n\cdot e^{-t(1+i)}dt=\frac{1}{n^{n+1}}\cdot\sum_{i=0}^{\infty}\left(\frac{1}{1+i}\right)^{n+1}\int_{0}^{\infty}s^ne^{-s}ds=\frac{1}{n^{n+1}}\cdot \zeta(n+1)\cdot\Gamma(n+1)$$ Using $\Gamma(n+1)=n\cdot\Gamma(n)$, we can conclude: $$\int_0^\infty \frac{x^n}{e^{nx}-1}dx=\frac{1}{n^n}\cdot\zeta(1+n)\cdot\Gamma(n)$$

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  • $\begingroup$ Though the idea being highlighted in comments, thanks for outlining the steps. $\endgroup$ Commented Aug 14, 2023 at 12:51

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