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According to WolframAlpha, the integral $$\int_0^{\infty} \log(x)^2 e^{-x} \, \mathrm{d}x$$ has closed form $\gamma^2 + \frac{\pi^2}{6}$, where $\gamma$ is the Euler-Mascheroni constant.

The term $\frac{\pi^2}{6}$ is $\zeta(2)$, and other integrals of the form $\int_0^{\infty} \log(x)^n e^{-x} \, \mathrm{d}x$ clearly have some relation to values of $\zeta(s)$ at integers. I do not know where to go with this. The usual trick I know for integrating something of the form $\log(x)^n f(x)$ is to compare integrals along the top and bottom of the slit in a keyhole contour, but this doesn't converge with the term $e^{-x}$.

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Using $$\Gamma(z) = \int_0^\infty e^{-x}x^{z-1} dx$$ we obtain $$\int_0^\infty e^{-x}\ln^2 x dx = \Gamma''(1) = \gamma^2+\zeta(2)$$


To derive $\Gamma''(1)$, use the expansion and take expoential on both sides: $$\ln\Gamma(1+z) = -\gamma z + \sum_{k=2}^\infty \frac{(-1)^k\zeta(k)}{k}z^k$$

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Let's use the classic differentiation under the integral sign to prove your identity. The work is as follows$$\begin{align*}\int\limits_0^{\infty}e^{-x}\log^2x\, dx & =\lim\limits_{\mu\to0}\frac {\partial^2}{\partial\mu^2}\int\limits_{0}^{\infty}e^{-x}x^\mu\, dx\\ & =\lim\limits_{\mu\to0}\frac {\partial^2\,\Gamma(\mu+1)}{\partial\mu^2}\\ & =\Gamma''(1)\end{align*}$$We can evaluate the gamma function by making use of the identity $\Gamma'(x)=\Gamma(x)\psi(x)$. Letting $x=1$ and differentiating, we see that$$\begin{align*}\Gamma''(1) & =\Gamma(1){\psi^2(1)}+\Gamma(1){\psi'(1)}\\ & ={\gamma^2}+{\zeta(2)}\end{align*}$$Hence, we can see immediately that$$\int\limits_{0}^{\infty}e^{-x}\log^2x\, dx=\gamma^2+\frac {\pi^2}6$$


You can generate similar identities by changing the power of the natural log function$$\begin{align*}\int\limits_0^{\infty}e^{-x}\log x\, dx & =\color{blue}{-\gamma}\\\int\limits_0^{\infty}e^{-x}\log^3 x\, dx & =\color{blue}{\gamma^3+\frac {\gamma\pi^2}2+2\zeta(3)}\\\int\limits_{0}^{\infty}e^{-x}\log^4 x\, dx & =\color{blue}{8\gamma\zeta(3)+\gamma^4+\gamma^2\pi^2+\frac {3\pi^4}{20}}\end{align*}$$

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We can make use of Gamma Function and differentiation of the integral $$ \int_0^{\infty} x^a e^{-x} d x=\Gamma(a+1) \tag*{(*)} $$ Differentiating both sides of $(*)$ at $a=0$ gives $$ \boxed{\left.\int_0^{\infty}(\ln x)^n e^{-x} d x=\frac{\partial^n}{\partial a^n}[\Gamma(a+1)] \right|_{ a=0}=\Gamma^{(n)}(1)} $$ For examples, $$ \int_0^{\infty} \ln x \cdot e^{-x} d x=\Gamma^{\prime}(1)=-\gamma $$ $$ \int_0^{\infty}(\ln x)^2 \cdot e^{-x} d x=\Gamma^{\prime \prime}(1)=\gamma^2+\frac{\pi^2}{6} $$ $$ \int_0^{\infty}(\ln x)^3 e^{-x} d x=\Gamma^{(3)}(1)=-\gamma^3-\frac{\gamma \pi^2}{2}-2 \zeta(3) \\ \int_0^{\infty}(\ln x)^4 e^{-x} d x=\Gamma^{(4)}(1)=\gamma^4+\gamma^2 \pi^2+\frac{3 \pi^4}{20}+8 \gamma \zeta (3) $$

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