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It would be very helpful if the following definite integral or a similar one had an analytic solution:

$$\int\nolimits_{-\infty}^{\infty}\mathrm{sech}^2(x)\exp(-\alpha x^2)\,\mathrm dx,\qquad \alpha \geq 0$$

I have attacked this problem from several directions now, including contour integration (the Gaussian blows up along Re=0), differentiation under the sign (no luck), and interpreting it as the expected value when sampling with a certain distribution (because both of these functions may be interpreted that way easily).

To me, it seems like this could have a nice formula because both functions separately do, there is so much symmetry, and the integrand seems like just a bunch of exponentials to me. I just wanted to know if anyone had a compelling reason why this won't have a 'nice' analytic solution - or even what the intuition of some more experienced people is concerning my probability of success or direction of my efforts.

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"integrand seems like just a bunch of exponentials" - it's actually quite easy to come up with something whose closed form we don't know with just a bunch of exponentials... – J. M. May 11 '11 at 2:55
"there is so much symmetry" - I agree, your integral looks so well-behaved that even the simple-minded trapezoidal rule is sufficient for numerically evaluating it... – J. M. May 11 '11 at 4:13
$\sin(x)$ has a nice integral, $1/x$ also has a nice integral. There is so much symmetry (e.g., both are odd functions). However, $\sin(x)/x$ doesn't have a nice integral...?! – Fabian May 11 '11 at 5:13
By the way, your integral is easy to evaluate numerically (as pointed out by J.M.) or approximately for both $\alpha$ small or large. – Fabian May 11 '11 at 5:14
P.S. $\int_{-\infty}^{\infty}\frac{\sin(x)}{x}\,dx=\pi$? – Chris M May 11 '11 at 13:01
up vote 3 down vote accepted


$$\begin{eqnarray*} I(\alpha ) &:&=\int_{-\infty }^{+\infty }\left( \text{sech }\left( x\right) \right) ^{2}\exp (-\alpha x^{2})\ \mathrm{d}x \\ &=&\int_{-\infty }^{+\infty }\left( \cosh x\right) ^{-2}\exp (-\alpha x^{2})\ \mathrm{d}x\qquad (1) \\ &=&\int_{-\infty }^{+\infty }\frac{4}{\left( e^{x}+e^{-x}\right) ^{2}}e^{-\alpha x^{2}} \ \mathrm{d}x. \end{eqnarray*}$$

Wolfram Alpha cannot evaluate $I(\alpha )$ in terms of standard mathematical functions, so most likely there is no closed form for it. In SWP I got the following expansion:

enter image description here

which I rewrote as

$$I(\alpha )=\sum_{k=0}^{\infty }4\left( -1\right) ^{k+1}e^{\frac{\left( k+1\right) ^{2}% }{\alpha }}\sqrt{\pi }\left( -1+\mathrm{erf }\left( \frac{k+1}{\sqrt{\alpha }}% \right) \right) \frac{k+1}{\sqrt{\alpha }},\qquad (2)$$

where $\mathrm{erf }(x)$ is the error function

$$\mathrm{erf }\left( x\right) =\frac{2}{\sqrt{\pi }}\int_{0}^{x}e^{-t^{2}}dt.\qquad(3)$$

In terms of the complementary error function

$$\mathrm{erfc }(x)=1-\mathrm{erf }(x)=\frac{2}{\sqrt{\pi }}\int_{x}^{\infty }e^{-t^{2}}\mathrm{d}t\qquad (4)$$

it may be written as

$$I(\alpha )=\frac{4\sqrt{\pi }}{\sqrt{\alpha }}\sum_{k=0}^{\infty } \left( -1\right) ^{k}e^{\left( k+1\right) ^{2}/\alpha } \left( k+1\right)\ \mathrm{erfc }\left( \frac{1}{\sqrt{\alpha }}\left( k+1\right) \right) . \qquad (5)$$

For $\alpha =1/2$, both $(1)$ and $(5)$ give $I(1/2)\approx 1.5183$; for $\alpha =1$, $I(1)\approx 1.2874$; and for $\alpha =10$, $I(10)\approx 0.53494.$

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Hi Americo. Thank you for the respond to such an old post! I have to ask for clarification though. Maybe I'm just making a dumb mistake, but the expansion you provided doesn't seem to be working correctly. Did you type it correctly? I ask this because the pair of brackets seems kind of awkward as written. – Chris M Jul 12 '11 at 4:15
@Chris: Hi. I added formula $(2)$ as outputed by SWP and the deduction of $(5)$ from $(2)$. Also I simplified the brackets and parentheses. – Américo Tavares Jul 12 '11 at 9:21
@Chris: I added a "print screen" of the SWP computation. – Américo Tavares Jul 13 '11 at 10:47

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