Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
A chain of substitutions yields the equivalent form $$f(a/b)=16b\int_0^1\frac{u^{a+b-1}du}{(u^{2a}+1)(u^{2b}+1)}.$$ Although my complex analysis is rusty, when I try partial fractions (via residue theorem) on this I unfortunately get tons of trigonometric expressions that I don't want to handle. At least it appears $f({\Bbb Q})\subseteq {\Bbb Q}^{\rm rab}\oplus\pi{\Bbb Q}^{\rm rab}$ as conjectured anyway. (By ${\Bbb Q}^{\rm rab}$ I mean the maximal real abelian extension of $\Bbb Q$, or equivalently $\Bbb Q$ with all values in $\cos(\pi\Bbb Q)$ adjoined.) –  anon Sep 12 '13 at 17:17

3 Answers 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}

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
can you, please, specify where one can find that paper. Because it is a specific one. –  Caran-d'Ache Sep 19 '13 at 5:37
@Caran-d'Ache, I believe that the paper is "Evaluation of a class of definite integrals" by M.L. Glasser. It claims to examine a class of integrals which would include $\int_{-\infty}^\infty \operatorname{sech} x \operatorname{sech} a x \, dx$ as a special case, but then goes on to evaluate a different special case (where the integrand equals $e^{-\alpha x} \operatorname{sech} x \operatorname{sech}(x + a) \operatorname{sech}(x + b)$ with $a \not= b \not= 0$ and $\operatorname{Re} \alpha| < 3$) which is not at all applicable to the one we are interested in. –  Kyle Dec 9 '13 at 4:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.