I want to evaluate the integral
$$I = \int_{-\infty}^\infty \frac{x^2}{(\cosh x)^\alpha} dx,$$
where $\alpha>0$ is a constant. The Mathematica says that the answer is
$$\frac{2^{\alpha +2} \, _4F_3\left(\frac{\alpha }{2},\frac{\alpha
}{2},\frac{\alpha }{2},\alpha ;\frac{\alpha }{2}+1,\frac{\alpha
}{2}+1,\frac{\alpha }{2}+1;-1\right)}{\alpha ^3}.$$
However, what I want is a simpler final answer than the above form involving hypergeometric function. I expect that a simplification is possible, since analogous simplification was possible for a similar integral. (See my previous question Integral where Mathematica gives hypergeometric function)
Best Answer
Using only Mathematica:
$$\int_{-\infty }^{\infty } \frac{x^2}{\cosh ^{\alpha }(x)} \, dx=\\\int_{-\infty }^{\infty } 2^{\alpha } \left(e^{-x}+e^x\right)^{-\alpha } x^2 \, dx=\\\mathcal{L}_s\left[\int_{-\infty }^{\infty } \mathcal{L}_A^{-1}\left[2^{\alpha } \left(e^{-x}+A e^x\right)^{-\alpha } x^2\right](s) \, dx\right](1)=\\\mathcal{L}_s\left[\int_{-\infty }^{\infty } \frac{2^{\alpha } e^{-e^{-2 x} s-x \alpha } s^{-1+\alpha } x^2}{\Gamma (\alpha )} \, dx\right](1)=\mathcal{L}_s\left[\frac{2^{-3+\alpha } s^{-1+\frac{\alpha }{2}} \Gamma \left(\frac{\alpha }{2}\right) \log ^2(s)}{\Gamma (\alpha )}-\frac{2^{-2+\alpha } s^{-1+\frac{\alpha }{2}} \Gamma \left(\frac{\alpha }{2}\right) \log (s) \psi ^{(0)}\left(\frac{\alpha }{2}\right)}{\Gamma (\alpha )}+\frac{2^{-3+\alpha } s^{-1+\frac{\alpha }{2}} \Gamma \left(\frac{\alpha }{2}\right) \psi ^{(0)}\left(\frac{\alpha }{2}\right)^2}{\Gamma (\alpha )}+\frac{2^{-3+\alpha } s^{-1+\frac{\alpha }{2}} \Gamma \left(\frac{\alpha }{2}\right) \psi ^{(1)}\left(\frac{\alpha }{2}\right)}{\Gamma (\alpha )}\right](1)=\\\frac{2^{-2+\alpha } \Gamma \left(\frac{\alpha }{2}\right)^2 \psi ^{(1)}\left(\frac{\alpha }{2}\right)}{\Gamma (\alpha )}$$ $$=\frac{\operatorname B\left(\frac\alpha2,\frac12\right)\psi^{(1)}\left(\frac\alpha2\right)}2$$
where: $\psi ^{(1)}\left(\frac{\alpha }{2}\right)$ is PolyGamma function,
$\mathcal{L}_A^{-1}[f(A)](s)$ is InverseLaplaceTransform,
$\mathcal{L}_A[f(A)](s)$ is LaplaceTransform.
MMA code:
LaplaceTransform[ Integrate[ InverseLaplaceTransform[2^\[Alpha] (E^-x + A*E^x)^-\[Alpha] x^2, A, s] // PowerExpand, {x, -Infinity, Infinity}, Assumptions -> {\[Alpha] > 0, s > 0}], s, A] /. A -> 1 // FullSimplify