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I was having trouble with this integral

Prove $$\int_{-\infty}^{\infty} e^{2x}x^2 e^{-e^{x}}dx=\gamma^2 -2\gamma+\zeta(2)$$

Where $\gamma$ is the euler mascheroni constant. Let $u=e^x\rightarrow \ln(u)= x$ thus $du=e^x dx$

$$\int_{-\infty}^{\infty} e^{2x}x^2 e^{-e^{x}}dx= \int_{0}^{\infty} u\ln^2(u) e^{-u}du$$ Now $$\Gamma(n)=\int_0^\infty x^{n-1}e^xdx$$ $$\Gamma'(n)=\int_0^\infty x^{n-1}e^x\ln(x)dx$$ $$\Gamma''(n)=\int_0^\infty x^{n-1}e^x\ln^2(x)dx\Rightarrow \Gamma''(2)=\int_0^\infty x e^x\ln^2(x)dx$$

How do I proceed from here? Do I have to use $\Gamma'(n)=\psi(n)\Gamma(n)$ ? How do I solve this integral ?

Thank you for your time

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2 Answers 2

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$$\Gamma''(2)=\Gamma''(s+2)\Big|_{s=0}=\left((s+1)\Gamma(s+1)\right)''\Big|_{s=0}=\left(\Gamma(s+1)+(s+1)\Gamma'(s+1)\right)'\Big|_{s=0} \\ =2\Gamma'(s+1)+(s+1)\Gamma''(s+1)\Big|_{s=0} = -2\gamma + \Gamma''(1) \, .$$ Now $$\Psi(s+1)+\gamma = \sum_{n=1}^\infty \frac{1}{n}-\frac{1}{n+s} \\ \Rightarrow \quad \Psi'(s+1)=\frac{{\rm d}}{{\rm d}s} \frac{\Gamma'(s+1)}{\Gamma(s+1)}=\frac{\Gamma''(s+1)}{\Gamma(s+1)}-\frac{\Gamma'(s+1)^2}{\Gamma(s+1)^2} = \sum_{n=1}^\infty \frac{1}{(n+s)^2} \\ \stackrel{s=0}{\Rightarrow} \quad \Psi'(1)=\Gamma''(1)-\Gamma'(1)^2=\Gamma''(1)-\gamma^2=\zeta(2)$$ and so $$\Gamma''(2)=-2\gamma+\gamma^2+\zeta(2) \, .$$

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One typo: it should be $e^{-x}$ in your last three equations. Yes go ahead, your integral is $I=\Gamma"(2)$, which can be evaluated as $\psi(z)\Gamma(z)=\Gamma'(z)$ d.w.r.t. $z$ then $$\Gamma"(z)=\psi'(z)\Gamma(z)+\psi(z)\Gamma'(z)=\psi'(z)\Gamma(z)+\psi^2(z)\Gamma(z) \implies \psi(2)=1-\gamma$$ $$\psi(x)==\gamma+\sum_{n=0}^{\infty} \left( \frac{1}{n+1}-\frac{1}{n+z}\right).$$ and $$\psi'(z)=\sum_{n=0}^{\infty}\frac{1}{(n+z)^2} \implies \psi'(2)=\zeta(2)-1.$$ Finally, $$I=\Gamma"(2)=(1-\gamma)^2+\zeta(2)-1$$

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