Is it possible to do this integral?
$$\int_{-\infty}^{\infty} \operatorname{sech}^2(Bx)\operatorname{sech}^2(ax) dx,$$ where $B$ and $a$ are constants.
I tried using Mathematica, but no dice...
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This doesn't give a nice enough answer to your question. But hopefully can be extended to give a nice enough answer. Let us assume $a,b>0$. Let $$I(a,b) = \displaystyle \int_{-\infty}^{\infty} \text{sech}^2(ax) \text{sech}^2(bx)dx = 2 \underbrace{\int_0^{\infty} \text{sech}^2(ax) \text{sech}^2(bx)dx}_{J(a,b)}.$$ We have \begin{align} \text{sech}^2(ax) \text{sech}^2(bx) & = \dfrac{16}{\left(\exp(ax) + \exp(-ax) \right)^2\left(\exp(bx) + \exp(-bx) \right)^2}\\ & = \dfrac{16 \exp(-2(a+b)x)}{(1+\exp(-2ax))^2(1+\exp(-2bx))^2}\\ & = 16 \exp(-(a+b)x) \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} (1+m)(1+k)(-1)^{m+k}\exp(-2(ak+bm)x)\\ & = 16 \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} (1+m)(1+k)(-1)^{m+k}\exp(-2(a(k+1)+b(m+1))x) \end{align} Hence, $$I(a,b) = 16 \sum_{m,k=0}^{\infty} \dfrac{(-1)^{m+k}(1+m)(1+k)}{a(k+1)+b(m+1)}$$
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First numerator should be $16$ after we have – Jean-Sébastien May 05 '13 at 21:21
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@Jean-Sébastien Thanks. Updated. – May 05 '13 at 21:26
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Thank you very much for this. I was hoping, however, for there to be a nicer solution one could write down. An expression this complicated won't be very helpful in my calculation. – Bob Riley May 06 '13 at 02:40