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.

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

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

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.

share|improve this question
    
"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
2  
P.S. $\int_{-\infty}^{\infty}\frac{\sin(x)}{x}\,dx=\pi$? –  Chris M May 11 '11 at 13:01

1 Answer 1

up vote 3 down vote accepted

Let

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

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 $\alpha =10$, $I(10)\approx 0.53494.$

share|improve this answer
    
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

Your Answer

 
discard

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.