Is is possible to evaluate $\int \frac{1-\text{sech}(\pi x)}{x} \, dx$ to some reasonable form?

Question

Is it possible to evaluate this integral to some reasonable formula? $$\int \frac{1-\text{sech}(\pi x)}{x} \, dx$$

I used integration by parts and got this result:

$$\int \frac{1-\text{sech}(\pi x)}{x} \, dx=-\frac{2}{\pi} \int \frac{\tan ^{-1}\left(\tanh \left(\frac{\pi x}{2}\right)\right)}{x^2} \, dx+\log (x)-\frac{2 \tan ^{-1}\left(\tanh \left(\frac{\pi x}{2}\right)\right)}{\pi x}$$

Hoping for simplifying somehow $\tan ^{-1}\left(\tanh \left(\frac{\pi x}{2}\right)\right)$ but without any success.

Answer 1

One method is to use the series expansion for $\text{sech}(x)$ in the form $$ \text{sech}(x) = \sum_{n=0}^{\infty} \frac{E_{2n} \, x^{2n}}{(2n)!} \hspace{10mm} |x| < \frac{\pi}{2}.$$ This leads to \begin{align} I &= \int \frac{1 - \text{sech}(\pi x)}{x} \, dx \\ &= - \sum_{n=1}^{\infty} \frac{E_{2n} \, \pi^{2n}}{(2n)!} \, \int x^{2n-1} \, dx \\ &= - \sum_{n=1}^{\infty} \frac{E_{2n} \, (\pi \, x)^{2n}}{(2n) \, (2n)!} + c_{0} \end{align} for $|x| < \frac{1}{2}$.

Answer 2

Here is a solution with justified steps at the end:

$$\mathrm{\int\frac{1-sech(\pi x)}{x}dx=ln(x)-\int \frac{sech(\pi x)}{x}dx=ln(x)-2\int \frac{1}{1-\left(-e^{2\pi x}\right)}\frac{e^{\pi x}}{x}dx=ln(x)-2\int \frac{e^{\pi x}}{x} \sum_{n=0}^\infty \left(-e^{2\pi x}\right)^n=\sum_{n=0}^\infty(-1)^n\int\frac{e^{(2n+1)\pi x}}{x}dx= \boxed{\mathrm{ C+ln(x)-2\sum_{n=0}^\infty(-1)^n Ei((2 n+1) π x)}}, \left| -e^{2\pi x}\right|<1\implies \boxed{\mathrm{Re(x)<0}}}$$

1. Integrate the reciprocal function.
2. Use the main definition of the hyperbolic secant function and use partial fractions.
3. Use the Best Friend Geometric Series expansion
4. Rearrange and factor using algebra and integrating term by term.
5. Use the definition of the Exponential Integral function
6. Simplify and add constant of integration.
7. First please see What is the integral of $\frac1x$?. Note the ln(x) is not a typo. The absolute value bars would cause the complex numbers to be calculated wrongly. This means the negative signs must cancel for x$< 0$ in integration , for example: $\mathrm{a,b\ge0:ln(-b)-ln(-a)=ln\left(\frac {-b}{-a}\right)=ln(b)-ln(a)=ln|-b|-ln|-a|}$

For an example see this computation and this other computation for a sum representation of:

$$\mathrm{\int_{-1}^{-1-3i} \frac{1-sech(\pi x)}{x}dx= \frac{ln(5)+ln(2)}{2}+tan^{-1}(3)i-2i\pi -2\sum_{n=0}^\infty \left[(-1)^n\big(Ei((-1-3i)(2n+1)\pi )-Ei(-(2n+1)\pi )\big)\right]=1.12618…+1.25671…i}$$

Notice that the function is odd, so you do not necessarily need $\mathrm{Re(x)>0}$. Please correct me and give me feedback!

Answer 3

Using the sequence of substitutions $$\pi x=y \qquad y=2 \tanh ^{-1}(z) \qquad z=\sqrt w\implies \color{blue} {x=\frac{2 }{\pi }\tanh ^{-1}\left(\sqrt{w}\right)}$$

$$I= \int \frac{1 - \text{sech}(\pi x)}{x} \, dx= \int \frac{\sqrt{w}}{\left(1-w^2\right) \tanh ^{-1}\left(\sqrt{w}\right)}\,dw$$

$$\frac{\sqrt{w}}{\tanh ^{-1}\left(\sqrt{w}\right)}=1-\sum_{n=1}^\infty a_n \,w^n$$

Looking at sequence $A216272$ in $OEIS$ $$a_n=\frac 1 {2n-1}\sum _{m=0}^{2 n-1} \frac{2^{m+1} }{m+1} \binom{2 n-1}{m}\sum _{k=0}^{m+1} \frac{k!\, S_{k+m}^{(m)}\, \mathcal{S}_{m+1}^{(k)}}{(k+m)!}$$ where appear Stirling numbers of the first and second kinds. The first coefficients $a_n$ are $$\left\{\frac{1}{3},\frac{4}{45},\frac{44}{945},\frac{428}{14175},\frac{10196}{46777 5},\frac{10719068}{638512875},\frac{25865068}{1915538625},\frac{5472607916}{4884 62349375},\cdots\right\}$$ $$I=\frac{1}{2} \log \left(\frac{1+w}{1-w}\right)-\sum_{n=1}^\infty a_n \int \frac {w^n}{1-w^2}\,dw$$ with $$J_n=\int \frac {w^n}{1-w^2}\,dw=\frac{w^{n+1} }{n+1}\,\, _2F_1\left(1,\frac{n+1}{2};\frac{n+1}{2}+1;w^2\right)$$ and the reccurences are simply

$$J_{2n+1}=J_{2n-1}-\frac{w^{2n}}{2n}\qquad \text{with} \qquad J_1=-\frac{1}{2} \log \left(1-w^2\right)$$ $$J_{2n+2}=J_{2n}-\frac{w^{2n-1}}{2n-1}\qquad \text{with} \qquad J_0=\tanh ^{-1}(w)$$

Now, integrating from $w=0$ to $w=\frac 9{10}$ that is to say from $x=0$ to $x=\frac{2 }{\pi }\tanh ^{-1}\left(\frac{3}{\sqrt{10}}\right)$, numerical integration gives as a result $1.09693$.

Computing $$K_p=\frac{1}{2} \log \left(\frac{1+w}{1-w}\right)-\sum_{n=1}^\infty a_n\, J_n$$ the results are $$\left( \begin{array}{cc} p & K_p \\ 10 & 1.09969 \\ 20 & 1.09725 \\ 30 & 1.09698 \\ 40 & 1.09694 \\ 50 & 1.09693 \end{array} \right)$$

Answer 4

This is exact formula with (if I am not mistaken) globally convergent series:

$$\frac{1-\text{sech}(\pi x)}{x}=\frac{1}{x}-\frac{4}{\pi x} \sum _{n=0}^{\infty } \frac{(-1)^n (2 n+1)}{(2 n+1)^2+4 x^2}$$

So we have:

$$\int \frac{1-\text{sech}(\pi x)}{x} \, dx=\frac{2}{\pi } \sum _{n=0}^{\infty } \frac{(-1)^n (2 n+1) \log \left((2 n+1)^2+4 x^2\right)}{(2 n+1)^2}$$

If someone is interested how I found these results - I used some of formulas from this wolfram site page and from Partial fraction expansion section of Wikipedia page on Trigonometric functions.