Proving $\int_0^\infty t e^{2t} e^{-e^{-2t}} \mathrm{d}t = \frac{-\gamma}{4}$.

Question

I have been trying to prove that $$ \int_0^\infty t e^{2t} e^{-e^{-2t}} \mathrm{d}t = \frac{-\gamma}{4} \tag{1} $$

My attempt: \begin{align*} \mathcal{I} &= \int_0^\infty t e^{2t} e^{-e^{-2t}} \, \mathrm{d}t \\ &= \frac{-1}{4} \int_0^1 \frac{\ln(y)}{y^2} e^{-y} \, \mathrm{d}y && (e^{-2t} = y) \\ &= \frac{-1}{4} \int_0^1 \frac{\ln(y)}{y^2} \left( 1 - y + \frac{y^2}{2!} - \frac{y^3}{3!} + \cdots \right) \, \mathrm{d}y \\ &= \frac{-1}{4} \bigg[\int_0^1 \frac{\ln(y)}{y^2} \, \mathrm{d}y - \int_0^1 \frac{\ln(y)}{y} \, \mathrm{d}y + \int_0^1 \frac{\ln(y)}{2!} \, \mathrm{d}y - \cdots \, \bigg] \end{align*}

But the first integral in last equation doesn't seem to converge. Other than that I can evaluate all integrals using gamma function's

$$ \int_0^1 x^n \ln^m(x) \, \mathrm{d}x = \frac{(-1)^m m!}{(n+1)^{m+1}} $$ where $m,n \in \mathbb{N}$

Please help me in proving $(1)$.

I literally have no idea that how it can be equal to $-\gamma/4$.

Thank you very much!!

EDIT: Actually it was $e^{2t}$ but I mistakenly typed $e^{-2t}$ if it was $e^{-2t}$ then it would be not this much tough.

EDIT: As @conan mentioned in the comment that integral should be $$ \int_{-\infty}^\infty t e^{2t} e^{-e^{2t}} \mathrm{d}t $$ Which is giving the required answer.

Answer 1

I think there is a typo in the integral. I think the original integral was $$\Omega := \int_0^\infty t e^{-2t} e^{-e^{-2t}} \,dt.$$ You can try integrating by substituting $t \mapsto -\ln(t)/2$: \begin{align*} \Omega &= -\frac{1}{4}\int_0^1 \ln(t) e^{{-t}} \,dt\\ &= -\frac{1}{4}\underbrace{\int_0^\infty \ln(t) e^{-t}\,dt}_{I_1} + \frac{1}{4} \underbrace{\int_1^\infty \ln(t) e^{-t}\,dt}_{I_2}. \end{align*} $I_1$ is evaluated in here to be $-\gamma$. For $I_2$, you would apply integration by parts with $u = \ln(t)$ and $dv = e^{-t}\,dt$: \begin{align*} I_2 &= \lim_{R \to \infty} \ -e^{-t} \ln(t) \Bigg|_{t = 1}^R + \int_1^\infty \frac{e^{-t}}{t}\,dt\\ &\stackrel{t \mapsto -t}{=} \int_1^\infty \frac{e^{-t}}{t}\,dt\\ &= -\int_{-\infty}^{-1} \frac{e^{t}}{t}\,dt\\ &= -\mathrm{Ei}(-1), \end{align*} since $$\mathrm{Ei}(x) = \int_{-\infty}^x \frac{e^t}{t}\,dt.$$ We can simplify further to $-\mathrm{Ei}(-1) = \delta/e$ where $\delta$ is the Euler-Gompertz constant. Thus, our integral evaluates to $$\Omega = \frac{1}{4}\left(\gamma + \frac{\delta}{e} \right).$$

Answer 2

As @conan mentioned in his comment, I also beleive the integral is: $$I = \int_{-\infty}^{\infty} x e^{2x}e^{-e^{2x}}\mathrm d x$$

Use $t := e^{2x}$,

$$I = \frac14 \int_{0}^{\infty} \ln (t)e^{-t}\mathrm d t = \frac14 \Gamma' (1) = -\frac\gamma 4$$

Answer 3

Prove that \begin{eqnarray*} \int_0^\infty te^{-2t}e^{-e^{-2t}} \, dt&=&-\frac\gamma 4, \end{eqnarray*} where $\gamma$ is the Euler-Mascheroni constant.

Solution. Substitute $x=e^t$, so we have \begin{eqnarray*} \int_0^\infty te^{-2t}e^{-e^{-2t}} \, dt&=& \int_0^\infty \frac{\log x}{x^2}e^{-1/x^2} \, \frac{dx}x \\&=&\int_0^\infty \frac 1{x^3}e^{-1/x^2} \log x\, dx \end{eqnarray*} Now, substitute $\displaystyle\frac 1x$ by $x$, we get \begin{eqnarray*} \int_0^\infty \frac 1{x^3}e^{-1/x^2} \log x\, dx&=&\int_0^\infty xe^{-x^2} \log x\,dx \end{eqnarray*} Finally, the substitution $x^2$ by $x$ yields \begin{eqnarray*} \int_0^\infty xe^{-x^2} \log x\,dx%&=&\left[-\frac 12e^{-x^2} \log x\right]_0^\infty+\int_0^\infty \frac{e^{-x^2}}x\,dx\\ &=& \frac 14\int_0^\infty e^{-x} \log x\,dx\\ &=& \frac 14\lim_{a\to 0}\frac{\partial}{\partial a}\left(\int_0^\infty x^ae^{-x}\,dx\right)\\ &=& \frac 14\lim_{a\to 0}\frac{\partial}{\partial a}\left(\Gamma(a+1)\right)\\ &=& \frac 14\lim_{a\to 0}\left(\Gamma'(a+1)\right)\\ &=& \frac 14\Gamma'(1)\\ &=& \frac 14\psi(1)\\ &=& -\frac\gamma 4, \end{eqnarray*} where $\psi(z)$ is a first derivative of $\log(\Gamma(z))$ function defined as: $$\psi(z)=\frac{\partial}{\partial z}(\log(\Gamma(z)))=\frac{\Gamma'(z)}{\Gamma(z)}=-\gamma+\sum_{n=1}^\infty\left(\frac 1n-\frac 1{n-1+z}\right).$$