The integral I'm trying to solve is
\begin{align*}
I = \int_{0}^{\infty} \log^2(t) t^{\frac{1}{2}} e^{-t} \, d t.
\end{align*}
Using Frullani's integral, note that
\begin{align*}
\log(t) = – \, \int_{0}^{\infty} \frac{e^{-xt}-e^{-x}}{x} \, d x.
\end{align*}
Hence,
\begin{align*}
I &= \int_{0}^{\infty} \log^2(t) t^{\frac{1}{2}} e^{-t} \, d t\\
&= \int_{0}^{\infty} \left[ \int_{0}^{\infty} \frac{e^{-xt}-e^{-x}}{x} \, d x \right] \left[ \int_{0}^{\infty} \frac{e^{-yt}-e^{-y}}{y} \, d y \right] t^{\frac{1}{2}} e^{-t} \, d t\\
&= \int_{0}^{\infty} \frac{1}{y} \int_{0}^{\infty} \frac{1}{x} \left[ \int_{0}^{\infty} t^{\frac{1}{2}} e^{-t} \left(e^{-xt}-e^{-x}\right) \left(e^{-yt}-e^{-y}\right) \, d t \right] \, d x \, d y\\
&= \int_{0}^{\infty} \frac{1}{y} \int_{0}^{\infty} \frac{1}{x} \left[ \int_{0}^{\infty} t^{\frac{1}{2}} e^{-t(1+x+y)} \, d t – \int_{0}^{\infty} t^{\frac{1}{2}} e^{-t(1+x)-y} \, d t \right.\\
&\qquad \left. – \, \int_{0}^{\infty} t^{\frac{1}{2}} e^{-t(1+y)-x} \, d t + \int_{0}^{\infty} t^{\frac{1}{2}} e^{-x-y-t} \, d t \right] \, d x \, d y\\
&= \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \frac{1}{y} \int_{0}^{\infty} \frac{1}{x} \left[ \frac{1}{(1+x+y)^{3/2}} – \frac{e^{-y}}{(1+x)^{3/2}} – \frac{e^{-x}}{(1+y)^{3/2}} + e^{-x-y} \right] \, d x \, d y.
\end{align*}
Integrating by parts with $\ d v=\frac{1}{x} \, d x$, we have
\begin{align*}
I &= \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \frac{1}{y} \left\{ \left[ \log(x) \left( \frac{1}{(1+x+y)^{3/2}} – \frac{e^{-y}}{(1+x)^{3/2}} – \frac{e^{-x}}{(1+y)^{3/2}} + e^{-x-y} \right) \right]_{x=0}^{x=\infty} \right.\\
&\qquad \left. – \, \int_{0}^{\infty} \left[ \frac{3e^{-y} \log(x)}{2(1+x)^{5/2}} + \frac{e^{-x} \log(x)}{(1+y)^{3/2}} – \frac{3\log(x)}{2(1+x+y)^{5/2}} – e^{-x-y} \log(x) \right] \, d x \right\} \, d y\\
&= – \, \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \frac{1}{y} \int_{0}^{\infty} \left[ \frac{3e^{-y} \log(x)}{2(1+x)^{5/2}} + \frac{e^{-x} \log(x)}{(1+y)^{3/2}} – \frac{3\log(x)}{2(1+x+y)^{5/2}} – e^{-x-y} \log(x) \right] \, d x \, d y\\
&= – \, \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \frac{1}{y} \left\{ \frac{3e^{-y}}{2} \int_{0}^{\infty} \frac{\log(x)}{(1+x)^{5/2}} – \frac{1}{(1+y)^{3/2}} \left[ – \int_{0}^{\infty} e^{-x} \log(x) \, d x \right] \right.\\
&\qquad \left. – \, \frac{3}{2} \int_{0}^{\infty} \frac{\log(x)}{(1+x+y)^{5/2}} \, d x + e^{-y} \left[ – \, \int_{0}^{\infty} e^{-x} \log(x) \, d x \right] \right\} \, d y\\
&= – \, \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \frac{1}{y} \left\{ 2e^{-y} [\log(2)-1] – \frac{1}{(1+y)^{3/2}} \gamma – \frac{\log(y+1)}{(y+1)^{3/2}} – \frac{2[\log(2)-1]}{(y+1)^{3/2}} + e^{-y} \gamma \right\} \, d y.
\end{align*}
Integrating by parts once again with $\ d v=\frac{1}{y} \, d y$, we have
\begin{align}
\nonumber I &= – \, \frac{\sqrt{\pi}}{2} \left\{ \left[ \log(y) \left( 2e^{-y} [\log(2)-1] – \frac{1}{(1+y)^{3/2}} \gamma – \frac{\log(y+1)}{(y+1)^{3/2}} \right. \right. \right.\\
\nonumber &\qquad \left. \left. \left. – \, \frac{2[\log(2)-1]}{(y+1)^{3/2}} + e^{-y} \gamma \right) \right]_{0}^{\infty} – \int_{0}^{\infty} \left[ \frac{3\log(y)}{2(1+y)^{5/2}} + \frac{3[\log(2)-1] \log(y)}{(y+1)^{5/2}} \right. \right.\\
\nonumber &\qquad \left. \left. – \, 2e^{-y} \log(y) [\log(2)-1] – \frac{\log(y)}{(y+1)^{5/2}} + \frac{3 \log(y+1) \log(y)}{2(y+1)^{5/2}} – e^{-y} \log(y) \gamma \right] \, d y \right\}\\
&= \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \left[ \frac{3\log(y)}{2(1+y)^{5/2}} + \frac{3[\log(2)-1] \log(y)}{(y+1)^{5/2}} – 2e^{-y} \log(y) [\log(2)-1] \right. \\
\nonumber &\qquad \left. – \, \frac{\log(y)}{(y+1)^{5/2}} + \frac{3 \log(y+1) \log(y)}{2(y+1)^{5/2}} – e^{-y} \log(y) \gamma \right] \, d y \\
\nonumber &= \ldots\\
\nonumber &= \frac{\sqrt{\pi}}{2} \left[4\log^2(2) + (4\gamma-8)\log(2) + \frac{\pi^2}{2} + \gamma^2 – 4\gamma\right].
\end{align}
I could almost solve the integral in the third-to-last line. However, in the said integral, I haven't found the answer to
\begin{align*}
\int_{0}^{\infty} \frac{3 \log(y+1) \log(y)}{2(y+1)^{5/2}} \, d y.
\end{align*}
Firstly, I have no idea how to solve this particular integral. Secondly, I'm also not sure if my whole calculation is already correct. Can anyone help me with this problem? I am open to any suggestions/solutions for solving the very last integral, or even for the very first integral in the question. Any help/feedback is much appreciated. Thank you!
Best Answer
$$ I = \int \log^2(t) \,\sqrt t\, e^{-t} \, dt=8\int \log ^2(x)\,x^2\,e^{-x^2} \,dx $$ can be computed in terms of hypergeometric functions but, I suppose that you are not interested by such an approach (it requires one integration by parts making the error function appearing in it).
With regard to $$J=\frac 3 2 \int \frac{\log (y)\, \log (y+1)}{(y+1)^{5/2}}\, dy$$ l $$y=x^2-1 \quad \implies \quad J=6 \int \frac{\log (x) \,\log \left(x^2-1\right)}{x^4}\,dx$$ One integration by parts $$J=-\frac{2 \log (x) \log \left(x^2-1\right)}{x^3}+\int \Bigg(\frac{4 \log (x)}{x^2 \left(x^2-1\right)}+\frac{2 \log \left(x^2-1\right)}{x^4} \Bigg)\,dx$$ which is more simple. Using a few more integrations by parts $$\int\frac{\log (x)}{x^2 \left(x^2-1\right)}\,dx=\frac{1}{x}-\frac{1}{2} \log (x+1) \log (x)+\frac{\log (x)}{x}-\frac{1}{2} (\text{Li}_2(1-x)+\text{Li}_2(-x))$$ $$\int\frac{\log \left(x^2-1\right)}{x^4} \,dx=-\frac{1}{3} \left(\frac{\log \left(x^2-1\right)}{x^3}-\frac{2}{x}+2 \tanh ^{-1}(x)\right)$$ Combining all the above and using the bounds \begin{align*} \int_{0}^{\infty} \frac{3 \log(y+1) \log(y)}{2(y+1)^{5/2}} \, d y=\frac{\pi ^2}2-\frac{4}{3} (4-\log (2)) \end{align*}