Integral $\int_{-\infty}^\infty \frac{x^2}{(\cosh x)^\alpha} dx$ for $\alpha>0$

Question

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)

Answer 1

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

Answer 2

Just to add the nice solution by @Mariusz Iwaniuk - we can find a general formula for $x^n (\,n=0,1,2...)$ in the integrand.

Let's denote $\displaystyle J(\beta)=\int_{-\infty}^\infty\frac{e^{\beta x}}{\cosh ^\alpha x}dx, \,\text{then} \,\,I(n)=\int_{-\infty}^\infty\frac{x^n}{\cosh ^\alpha x}dx=\frac{\partial^n}{\partial\beta^n}J(\beta)\bigg|_{\beta=0}$

and $$J(\beta)=2^\alpha\int_{-\infty}^\infty\frac{e^{(\alpha+\beta) x}}{(e^{2x}+1) ^\alpha }dx\overset{t=e^x}=2^\alpha\int_0^\infty\frac{t^{\alpha+\beta-1}}{(t^2+1) ^\alpha }dt=2^{\alpha-1}\int_0^\infty\frac{x^{\frac{\alpha+\beta}{2}-1}}{(x+1) ^\alpha }dx$$ $$\overset{t=\frac{1}{x+1}}{=}\,2^{\alpha-1}\int_0^1(1-t)^{\frac{\alpha+\beta}{2}-1}t^{\frac{\alpha-\beta}{2}-1}dt=2^{\alpha-1} B\Big(\frac{\alpha+\beta}{2};\frac{\alpha-\beta}{2}\Big)=\frac{2^{\alpha-1}\Gamma\big(\frac{\alpha+\beta}{2}\big)\Gamma\big(\frac{\alpha-\beta}{2}\big)}{\Gamma(\alpha)}$$

$$\boxed{\,I(n)=\frac{2^{\alpha-1}}{\Gamma(\alpha)}\frac{\partial^n}{\partial\beta^n}e^{\ln\Gamma\big(\frac{\alpha+\beta}{2}\big)+\ln\Gamma\big(\frac{\alpha-\beta}{2}\big)}\bigg|_{\beta=0}=\frac{2^{\alpha-1-n}}{\Gamma(\alpha)}\frac{\partial^n}{\partial s^n}e^{\ln\Gamma\big(\frac{\alpha}{2}+s\big)+\ln\Gamma\big(\frac{\alpha}{2}-s\big)}\bigg|_{s=0}\,}$$ For the particular cases $n=0,\,1,\,2,...$ $$I(0)=\frac{2^{\alpha-1}\Gamma^2\big(\frac{\alpha}{2}\big)}{\Gamma(\alpha)}$$ $$I(1)=\frac{2^{\alpha-2}\Gamma\big(\frac{\alpha}{2}+s\big)\Gamma\big(\frac{\alpha}{2}-s\big)}{\Gamma(\alpha)}\Big(\psi\big(\frac{\alpha}{2}+s\big)-\psi\big(\frac{\alpha}{2}-s\big)\Big)\bigg|_{s=0}=0$$ $$I(2)=\frac{2^{\alpha-3}\Gamma^2\big(\frac{\alpha}{2}\big)}{\Gamma(\alpha)}\bigg[\Big(\psi\big(\frac{\alpha}{2}\big)-\psi\big(\frac{\alpha}{2}\big)\Big)^2+\Big(\psi^{(1)}\big(\frac{\alpha}{2}\big)+\psi^{(1)}\big(\frac{\alpha}{2}\big)\Big)\bigg]=\frac{2^{\alpha-2}\Gamma^2\big(\frac{\alpha}{2}\big)}{\Gamma(\alpha)}\psi^{(1)}\Big(\frac{\alpha}{2}\Big)$$ etc.