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I'm looking for a proof of this:

$$\int\limits_0^\infty \mathcal{e}^{-x}\ln^{2}x\,dx = \gamma^{2}+\frac{\pi^{2}}{6}$$

My first thought would have been to write $e^{-x}$ as an infinite series and then exchange its order with the integral, but that brought me nowhere. Considering also that $\frac{\pi^{2}}{6}$ shows up, it seems inevitable to make use of the zeta function, which is something I'm not really familiar with. For context, I'm trying to evaluate the Laplace Transform of $\ln^2x$, which has come down to the integral above.

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    $\begingroup$ Hint: it is the second derivative of the $\Gamma $-function at $z=1$. $\endgroup$ – user Mar 26 at 23:28
  • $\begingroup$ @marcozz Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Apr 3 at 3:22
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Recall that $$\Pi(s)=\Gamma(s+1)=\int_0^\infty x^{s}e^{-x}dx$$ so by the Leibniz integral rule, $$\Pi'(s)=\int_0^\infty \frac{\partial}{\partial s}x^se^{-x}dx=\int_0^\infty x^{s}e^{-x}\ln(x)dx$$ So naturally $$\Pi''(s)=\int_0^\infty x^se^{-x}\ln(x)^2dx$$ so your integral is given by $$\Pi''(0)=\Gamma''(1)$$ then we recall the definition of the polygamma function: $$\psi_n(s)=\left(\frac{d}{ds}\right)^{n+1} \ln\Gamma(s)=\frac{d}{ds}\psi_{n-1}(s)$$ So we see that $$\Gamma'(s)=\Gamma(s)\psi_0(s)$$ And accordingly, $$\Gamma''(s)=\Gamma'(s)\psi_0(s)+\Gamma(s)\psi_1(s)$$ Which is $$\Gamma''(s)=\Gamma(s)\left(\psi_0^2(s)+\psi_1(s)\right)$$ Hence your integral is $$\Gamma''(1)=\psi_0^2(1)+\psi_1(1)$$ Then from here we have that $$\psi_0(1)=-\gamma$$ And from here $$\psi_1(1)=\frac{\pi^2}6$$ So we have your integral at $$\Gamma''(1)=\gamma^2+\frac{\pi^2}6$$

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For any $x>0$, we can write

$$\log(x)=\int_0^\infty \frac{e^{-t}-e^{-xt}}{t}\,\,dt\tag1$$

Using $(1)$, we have

$$\begin{align} \int_0^\infty e^{-x}\log^2(x)\,dx&=\int_0^\infty \int_0^\infty \frac{e^{-(t+x)}-e^{-x(t+1)}}{t}\log(x)\,dt\,dx\\\\ &=\int_0^\infty\frac1t \int_0^\infty \left(e^{-(t+x)}-e^{-x(t+1)}\right)\log(x)\,dx\,dt\\\\ &=\int_0^\infty \frac1t\left(-\gamma e^{-t}+\frac\gamma{t+1}\right)\,dt-\int_0^\infty \frac{\log(t+1)}{t(t+1)}\tag2 \end{align}$$


Integrating by parts the first integral on the right-hand side of $(2)$ with $u=\left(-\gamma e^{-t}+\frac\gamma{t+1}\right)$ and $v=\log(t)$, we obtain

$$\begin{align} \int_0^\infty \frac1t\left(-\gamma e^{-t}+\frac\gamma{t+1}\right)\,dt&=-\int_0^\infty \left(\gamma e^{-t}-\frac\gamma{(t+1)^2}\right)\log(t)\,dt\\\\ &=\gamma^2+\gamma\underbrace{\int_0^\infty \frac{\log(t)}{(t+1)^2}\,dt}_{=0\,\text{as seen by substituting }\,t\mapsto 1/t}\\\\ &=\gamma^2 \tag3 \end{align}$$


For the second integral on the right-hand side of $(2)$, we have

$$\begin{align} \int_0^\infty \frac{\log(t+1)}{t(t+1)}\,dt&=\int_0^1 \frac{\log(t+1)}{t(t+1)}\,dt+\int_1^\infty \frac{\log(t+1)}{t(t+1)}\,dt\\\\ &=\int_0^1 \left(\frac{\log(t+1)}{t}-\frac{\log(t+1)}{t+1}\right)\,dt+\int_0^1 \frac{\log(t+1)-\log(t)}{t+1}\,dt\\\\ &\int_0^1 \left(\frac{\log(t+1)}{t}\right)\,dt-\int_0^1 \left(\frac{\log(t)}{t+1}\right)\,dt\\\\ &=2\int_0^1 \left(\frac{\log(t+1)}{t}\right)\,dt\\\\ &=2\frac{\pi^2}{12}\\\\ &=\frac{\pi^2}{6}\tag4 \end{align}$$


Putting it all together yields the coveted identity

$$\int_0^\infty e^{-x}\log^2(x)\,dx=\gamma^2+\frac{\pi^2}{6}$$

as was to be shown!

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  • 2
    $\begingroup$ "thou shalt not covet thy neighbor's identity"... +1 $\endgroup$ – clathratus Mar 27 at 1:07
  • $\begingroup$ @clathratus Thank you!! Much appreciated. $\endgroup$ – Mark Viola Mar 27 at 1:09
  • $\begingroup$ @marcozz Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Apr 3 at 3:21

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