Background
Hi. I am currently writing my undergraduate thesis which mainly revolves around the generalized log-Moyal distribution pioneered by Bhati and Ravi (see here). In the aforementioned article, Bhati and Ravi addressed a proposition regarding the generalized log-Moyal distribution, one of which is the following:
$$E\left[\log\left(\frac{\mu}{Y}\right)\right] = -\sigma(\log 2+\gamma),$$
where $\gamma=-\int_{0}^{\infty}\log(t)\exp\{-t\}dt\approx 0.577216$ is Euler's gamma constant. The two authors did not provide the proof for this proposition. However, they mentioned that the proof is followed by straightforward computations. I have attempted to prove this proposition, but I have been stuck for quite some time trying to complete the third-to-last line of the following proof.
Attempted Proof
Notice that
\begin{align*}
E\left[\log\left(\frac{\mu}{Y}\right)\right] &= \int_{0}^{\infty} \log\left(\frac{\mu}{y}\right) f(y) d y\\
&= \int_{0}^{\infty} \log\left(\frac{\mu}{y}\right) \frac{1}{\sqrt{2\pi} \sigma y} \left(\frac{\mu}{y}\right)^{1/(2\sigma)} \exp\left\{-\frac{1}{2}\left(\frac{\mu}{y}\right)^{1/\sigma}\right\} d y\\
&= \frac{1}{\sqrt{2\pi} \sigma} \int_{0}^{\infty} \log\left(\frac{\mu}{y}\right) y^{-1-\frac{1}{2\sigma}} \mu^{\frac{1}{2\sigma}} \exp\left\{-\frac{1}{2}\left(\frac{\mu}{y}\right)^{1/\sigma}\right\} d y.
\end{align*}
Applying the substitution $t=\frac{1}{2}\left(\frac{\mu}{y}\right)^{1/\sigma}$, we have $y=\frac{\mu}{(2t)^{\sigma}}$ such that $d y = \frac{\mu}{2^{\sigma}}(-\sigma)t^{-\sigma-1}d t$. Hence, $t\to \infty$ as $y\to 0$ and $t\to 0$ as $y\to \infty$, implying that
\begin{align*}
E\left[\log\left(\frac{\mu}{Y}\right)\right] &= \frac{1}{\sqrt{2\pi} \sigma} \int_{\infty}^{0} \log\left[\frac{\mu}{\mu/(2t)^{\sigma}}\right] \left[\frac{\mu}{(2t)^{\sigma}}\right]^{-1-\frac{1}{2\sigma}} \mu^{\frac{1}{2\sigma}} \exp\left\{-t\right\} \left[\frac{\mu}{2^{\sigma}} (-\sigma) t^{-\sigma-1}\right] d t\\
&= -\frac{1}{\sqrt{2\pi} \sigma} \int_{0}^{\infty} \log\left[(2t)^{\sigma}\right] \left[\frac{(2t)^{\sigma}}{\mu}\right]^{1+\frac{1}{2\sigma}} \mu^{\frac{1}{2\sigma}} \exp\left\{-t\right\} \left[\frac{\mu}{2^{\sigma}} (-\sigma) t^{-\sigma-1}\right] d t\\
&= -\frac{1}{\sqrt{2\pi} \sigma} \int_{0}^{\infty} \log\left[(2t)^{\sigma}\right] \frac{(2t)^{\sigma\left(1+\frac{1}{2\sigma}\right)}}{\mu^{1+\frac{1}{2\sigma}}} \mu^{\frac{1}{2\sigma}} \exp\left\{-t\right\} \left[\frac{\mu}{2^{\sigma}} (-\sigma) t^{-\sigma-1}\right] d t\\
&= -\frac{1}{\sqrt{2\pi} \sigma} \int_{0}^{\infty} \log\left[(2t)^{\sigma}\right] \frac{2^{\sigma+\frac{1}{2}} \cdot t^{\sigma+\frac{1}{2}}}{\mu^{1+\frac{1}{2\sigma}}} \mu^{\frac{1}{2\sigma}} \exp\left\{-t\right\} \left[\frac{\mu}{2^{\sigma}} (-\sigma) t^{-\sigma-1}\right] d t\\
&= \frac{-(-\sigma)2^{\sigma+\frac{1}{2}}\cdot\mu^{\frac{1}{2\sigma}}\cdot\mu}{\sqrt{2\pi}\sigma \cdot 2^{\sigma} \cdot \mu^{1+\frac{1}{2\sigma}}} \int_{0}^{\infty} \log\left[(2t)^{\sigma}\right] t^{\sigma+\frac{1}{2}} \cdot t^{-\sigma-1} \exp\{-t\} d t\\
&= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \log\left[(2t)^{\sigma}\right] t^{\frac{1}{2}-1} \exp\{-t\} d t\\
&= \frac{\sigma}{\sqrt{\pi}} \int_{0}^{\infty} \left[\log(2)+\log(t)\right] t^{\frac{1}{2}-1} \exp\{-t\} d t\\
&= \frac{\sigma}{\sqrt{\pi}} \left[\log(2) \int_{0}^{\infty} t^{\frac{1}{2}-1} \exp\{-t\} d t + \int_{0}^{\infty} \log(t) t^{\frac{1}{2}-1} \exp\{-t\} d t\right]\\
&= \frac{\sigma}{\sqrt{\pi}} \left[\log(2) \Gamma\left(\frac{1}{2}\right) + \int_{0}^{\infty} \log(t) t^{\frac{1}{2}-1} \exp\{-t\} d t\right]\\
&= \frac{\sigma}{\sqrt{\pi}} \left[\log(2) \sqrt{\pi} + \int_{0}^{\infty} \log(t) t^{\frac{1}{2}-1} \exp\{-t\} d t\right]\\
&= \cdots\\
&= -\sigma\left[\log 2-\int_{0}^{\infty} \log(t) \exp\{-t\} d t\right]\\
&= -\sigma(\log 2+\gamma).
\end{align*}
Question
So, my main question: how do you prove that
\begin{align*}
\int_{0}^{\infty} \log(t) t^{\frac{1}{2}-1} \exp\{-t\} d t &= -\sqrt\pi\left[2\log(2)-\int_{0}^{\infty} \log(t) \exp\{-t\} d t\right]\\
&= -\sqrt\pi[2\log(2)+\gamma].
\end{align*}
I have tried a handful of combinations for the substitution method, but I am starting to doubt that the process of achieving the second-to-last line of my proof is not as easy as I thought it was (or maybe it is, but I overlooked it?). I have also read the proofs to this question asked by the Peruvian mathematician Harald Andrés Helfgott over 7 years ago. I would say that robjohn's proof is rigid, but unfortunately, it is too complicated for my thesis and myself as I mainly focus on actuarial sciences, and my thesis in particular revolves around loss models and risk theory. Can anybody help me with this challenge I've come across? Any help is much appreciated. Thanks!
Best Answer
Too long for a comment
Using the Frullani' integral $\,\,\displaystyle \ln (t)=-\int_0^\infty\frac{e^{-xt}-e^{-x}}xdx\,\,$ and changing the order of integration $$\int_0^\infty \ln (t) \,t^{-\frac12} e^{-t} dt=-\int_0^\infty\frac{dx}x\int_0^\infty\left(e^{-(x+1)t}-e^{-t}e^{-x}\right)t^{-\frac12}dt$$ $$=-\int_0^\infty\frac{dx}x\left(\frac1{\sqrt{1+x}}-e^{-x}\right)\int_0^\infty t^{-\frac12}e^{-t}dt=\sqrt\pi\int_0^\infty\left(e^{-x}-\frac1{\sqrt{1+x}}\right)\frac{dx}x$$ Integrating by parts, we have $$=\sqrt\pi\ln x\left(e^{-x}-\frac1{\sqrt{1+x}}\right)\,\bigg|_{x=0}^\infty-\sqrt\pi\int_0^\infty\left(\frac{\ln x}{2(1+x)^{\frac32}}-e^{-x}\ln x\right)dx$$ $$=-\sqrt\pi\,\gamma-\frac{\sqrt\pi}2\int_0^\infty\frac{\ln x}{(1+x)^{\frac32}}dx\overset{t=\frac1{\sqrt{1+x}}}{=}-\sqrt\pi\,\gamma-\sqrt\pi\int_0^1\ln \frac{1-t^2}{t^2}dt$$ $$=-\sqrt\pi\,\gamma+2\sqrt\pi\int_0^1\ln (t)\,dt-\sqrt\pi\int_0^1\ln(1-t)dt-\sqrt\pi\int_0^1\ln(1+t)dt$$ $$I=-\sqrt\pi\big(\gamma+2\ln2\big).$$