The integral : $\frac{1}{2}\int_0^\infty x^n \operatorname{sech}(x)\mathrm dx$

How can I evaluate

$$\frac{1}{2}\int_0^\infty x^n \operatorname{sech}(x)\mathrm dx?$$

I was trying integration by parts but it seemed like it is getting more complicated. $$\int_0^\infty x^n \operatorname{sech}(x)\mathrm dx=\left.2x^n\arctan\left(\tanh(x/2)\right)\right|_0^\infty-2n\int_0^\infty x^{n-1}\arctan(\tanh(x/2))\mathrm dx$$ Herein, it seems like we have to apply integration by parts $$n$$ times but it is not practically possible.

This question is a more general problem of the integral $$\int_0^\infty \frac{x}{e^x+e^{-x}}\mathrm dx$$, which I was first solving. Let me know if there's any other method for evaluating this integral. It will be highly appreciated.

I have posted my solution employing a method using Geometric series to which this Wikipedia article helped me in finding the solution. Please see my answer below.

• Is there anything wrong with this post? I can see a close vote here. Perhaps 'answer your own question' is controversial in practice as opposed to what was mentioned here.
– user730361
Commented May 25, 2021 at 5:58
• Or perhaps you should indicate some alternative approach in the question which had some issue and you later resolved the issue and posted that as an answer. Commented May 25, 2021 at 9:15
• @ParamanandSingh One user commented - 'Still the question should have a context' - that's why I added this. This question cannot be solved by integration by parts or if it can, it is going to be extremely tough.
– user730361
Commented May 25, 2021 at 9:21
• Also adding context just for the sake of formality is not really the spirit here. If someone asks you for a context you need to honestly describe how you stumbled on the problem and what are your own ideas for a solution and perhaps you need to write in such a way that it motivates other users to think about the problem. Commented May 25, 2021 at 9:28

\begin{align}\mathcal{I}&=\frac{1}{2}\int_0^\infty \frac{x^n}{\cosh(x)}\mathrm dx\\&=\int_0^\infty \frac{x^n}{e^x+e^{-x}}\mathrm dx\\&=\int_0^\infty \frac{x^ne^{-x}}{1+e^{-2x}}\mathrm dx.\end{align}

Proposition: $$\int_0^\infty \frac{x^{s-1}e^{-x}}{1+e^{-2x}}\mathrm dx=\beta(s)\Gamma(s),$$ where $$\beta(s)$$ is the Dirichlet beta function defined as $$\beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$$.

Proof:\begin{align}\int_0^\infty x^{s-1}\left(\frac{e^{-x}}{1-(-e^{-2x})}\right)\mathrm dx&=\int_0^\infty x^{s-1}\left(\sum_{k=0}^\infty (-1)^ke^{-(2k+1)x}\right)\mathrm dx \text{, using geometric series}\\&=\sum_{k=0}^\infty (-1)^k\int_0^\infty x^{s-1} e^{-(2k+1)x}\mathrm dx\\&=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^s}\displaystyle\int_0^\infty x^{s-1}e^{-x}\mathrm dx\text{, substituting (2k+1)x\mapsto x}\\&=\beta(s)\Gamma(s). \end{align}

Therefore, $$\mathcal{I}=\beta(n+1)\Gamma(n+1).$$

We can evaluate for different values of $$n\ge 0$$. For $$n=1$$, $$\frac{1}{2}\int_0^\infty x\operatorname{sech}(x)=\beta(2)\Gamma(2)=G$$, where $$G$$ is Catalan's constant. For $$n=2$$, $$\int_0^\infty x^2\operatorname{sech}(x)=2\beta(3)\Gamma(3)=\frac{\pi^3}{8}$$

• Why did you post your attempt as an answer rather in the main body of the OP? Commented May 25, 2021 at 5:33
• @MartinØdegaard Refer here.
– user730361
Commented May 25, 2021 at 5:35
• @meta, in my opinion, a self-answer obviates the need for context. Commented May 27, 2021 at 2:18
• Another opinion completely opposite to Gerry's one here Commented May 27, 2021 at 2:54
• @ArcticChar I've never seen anyone dispute in meta that self-answers should not count toward context. That a self-answer is context-in-a-different-location is sort of a no-brainer. Commented May 27, 2021 at 14:11

Alternatively

\begin{align} \frac{1}{2}\int_0^\infty \frac{x^n}{\cosh x}dx \overset{t=e^{-x}}=&\int_0^1 \frac{(-1)^n\ln^n t}{1+t^2}dt=n!\>\text{Im}\>\text{Li}_{n+1}(i) \end{align}

Letting $$\tan \theta=e^{-x}$$ transforms the integral into

$$\int_0^{\infty} \frac{x^n}{e^x+e^{-x}}dx=(-1)^n \int_0^{\frac{\pi}{4}} \ln^n (\tan \theta) d \theta$$

By my post, $$\int_{0}^{\frac{\pi}{4}} \ln ^{n}(\tan \theta) d \theta =(-1)^{n} \beta(n+1) \Gamma(n+1)$$ Hence$$\boxed{\int_0^{\infty} \frac{x^n}{e^x+e^{-x}}= \beta(n+1) \Gamma(n+1)}$$