Closed form for $\int_{-\infty}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx$

Question

Let $$f(a)=\int_{-\infty}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx,$$ where $\operatorname{sech}(z)=\frac2{e^z+e^{-z}}$ is the hyperbolic secant.

Here are values of $f(a)$ at some particular points: $$f(0)=\pi,\hspace{.15in}f(1)=2,\hspace{.15in}f(2)=\left(\sqrt2-1\right)\,\pi,\hspace{.15in}f\left(\frac34\right)=\left(4\sqrt{2+\sqrt2}-\frac{20}3\right)\,\pi.$$ Athough I do not yet have a proof ready, it seems that for every $a\in\mathbb{Q},\ f(a)=\alpha+\beta\,\pi$, where $\alpha$ and $\beta$ are algebraic numbers.

I wonder, if it is possible to express $f\left(\sqrt2\right)$ in a closed form?

Answer 1

Note that $af(a)=f(1/a)$. For similar integrals see Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz.76,498-No.541(1975), 49-50. It is amusing to note that $$\int_{-\infty}^{\infty}\operatorname{sech}(x) \operatorname{sech}[ax(x+i\pi)]\,\mathrm dx=\pi \operatorname{sech}(\pi^2 a/4)$$ but I doubt $f(a)$ has a closed form expression.

Answer 2

You can have this form of solution

$$ \int_{-\infty}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx =\frac{2}{a}\sum _{k=0}^{\infty } \left( -1 \right)^{k} \left( \psi \left( \,{\frac {3\,a+2\,k+1}{4a}} \right) -\psi \left( {\frac {2\,k+1+a} {4a}} \right) \right),$$

where $\psi(x)$ is the digamma function. Note that, $a=0$ is a special case.

Answer 3

${\large\mbox{We just need to evaluate}\ {\rm f}\left(a\right)\ \mbox{when}\ a \in \left\lbrack 0, 1\right\rbrack}$ since $$ {\rm f}\left(-a\right) = {\rm f}\left(a\right) \quad\mbox{and}\quad {\rm f}\left(1 \over a\right) = \left\vert a\right\vert\,{\rm f}\left(a\right) $$

\begin{align} {\rm f}\left(1 \over a\right) &= \int_{-\infty}^\infty {\rm sech}\left(x\right)\,{\rm sech}\left(x \over a\right)\,{\rm d}x = a\int_{-\infty}^\infty {\rm sech}\left(a\,{x \over a}\right){\rm sech}\left(x \over a\right) \,{{\rm d}x \over a} \\[3mm]&= \left\vert a\right\vert\int_{-\infty}^\infty {\rm sech}\left(ax\right)\,{\rm sech}\left(x\right)\,{\rm d}x = \left\vert a\right\vert\,{\rm f}\left(a\right) \end{align}

For example \begin{align} {\rm f}\left(1 \over 2\right) &= 2\,{\rm f}\left(2\right) = 2\left(\sqrt{2\,} - 1\right)\pi \\[3mm] {\rm f}\left(4 \over 3\right) &= {3 \over 4}\,{\rm f}\left(3 \over 4\right) = \left(3\sqrt{2 +\sqrt{\vphantom{\large A}2\,}\,} - 5\right)\,\pi \\[3mm] {\rm f}\left(a\right) & = {1 \over \left\vert a\right\vert}\,{\rm f}\left(1 \over a\right) \approx {1 \over \left\vert a\right\vert}\,{\rm f}\left(0\right) = {\pi \over \left\vert a\right\vert}\,, \quad \left\vert a \right\vert \gg 1 \end{align}

Answer 4

use descomposition as $$\sum _{k=0}^{\infty } \frac{4 i \text{a1} (-1)^{k+1} x \sec \left(\frac{1}{2} \pi \text{a1} (2 k+1)\right)}{\pi (2 k+1) (\pi (2 k+1)+2 i x)}+\sum _{k=0}^{\infty } \frac{4 i (-1)^{k+1} x \sec \left(\frac{1}{2} \pi \text{a1} (2 k+1)\right)}{\pi (2 k+1) (\pi (2 k+1)+2 i \text{a1} x)}+\sum _{k=0}^{\infty } \frac{4 i (-1)^k x \sec \left(\frac{1}{2} \pi \text{a1} (2 k+1)\right)}{\pi (2 k+1) (\pi (2 k+1)-2 i x)}+\sum _{k=0}^{\infty } \frac{4 i \text{a1} (-1)^k x \sec \left(\frac{1}{2} \pi \text{a1} (2 k+1)\right)}{\pi (2 k+1) (\pi (2 k+1)-2 i \text{a1} x)}=\text{sech}(x) \text{sech}(\text{a1} x)$$ and integrating term by term gives $$\sum _{k=0}^{\infty } \frac{\pi (\text{a1}+1) (-1)^k \sec \left(\pi \text{a1} k+\frac{\pi \text{a1}}{2}\right) \left(\text{a1}^2 (-1)^{\left\lfloor \frac{\arg (2 k+1)}{\pi }+\frac{1}{2}\right\rfloor }+(-1)^{\left\lfloor \frac{-2 \arg (\text{a1})+2 \arg (2 k+1)+\pi }{2 \pi }\right\rfloor }\right)}{\text{a1}^2}$$