# How do we prove this conjecture $\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}=2\gamma ?$

I was observing this question by @Brightsun and conjecture $(1)$

$$\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}=2\gamma \tag1$$

An attempt

$x=u^2$ then $(1)$ becomes

$$2\int_{0}^{\infty}ue^{-u}\ln{\left(1+{1\over u}\right)}\tag2$$

$$2\sum_{n=0}^{\infty}{(-1)^n\over n!}\int_{0}^{\infty}u^{n+1}\ln{\left(1+{1\over u}\right)}\mathrm du\tag3$$

Changing $(2)$ by applying $e^x$, $(3)$ diverges.

How would one prove $(1)?$

• You can't interchange the sum and integral like that because, as you noted, the integral diverges. You really need absolute convergence for that operation to be valid. Apr 10, 2017 at 5:32
• In general, $$n!~=~\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~dx,$$ and $$H_n~=~\int_0^1\ln\Big(1-\sqrt[n]x\Big)~dx.$$ Sep 2, 2017 at 16:29
• Furthermore, $$\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~\ln\Big(\sqrt[n]x\Big)~dx ~=~ n!~\big(H_{n-1}-\gamma\big).$$ Sep 2, 2017 at 16:41

By the change of variable $$u=\sqrt{x},\quad x=u^2, \quad dx=2udu,$$ one has \begin{align} \int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}dx&=2\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+{1\over u}\right)}du \\\\&=2\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+u\right)}\:du-2\int_{0}^{\infty}u\:e^{-u}\ln{u}\:du. \tag1 \end{align} By integrating by parts, the first integral on the right hand side of $(1)$ gives \begin{align} &\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+u\right)}\:du \\\\&=\left[\frac{e^{-u}}{-1}\cdot u\:\ln{\left(1+u\right)}\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-u}\left(\ln{\left(1+u\right)}+\frac{u}{1+u}\right)du \\\\&=\int_{0}^{\infty}e^{-u}\left(\ln{\left(1+u\right)}+\frac{u+1-1}{1+u}\right)du \\\\&=\int_{0}^{\infty}e^{-u}\left(\color{red}{\ln{\left(1+u\right)}-\frac{1}{1+u}}\right)du+\int_{0}^{\infty}e^{-u}du \\\\&=\color{red}{0}+1 \quad (\text{integration by parts}) \end{align} By integrating by parts, the second integral on the right hand side of $(1)$ gives \begin{align} &\int_{0}^{\infty}u\:e^{-u}\ln u\:du \\\\&=\left[\frac{e^{-u}}{-1}\cdot u\:\ln u\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-u}\left(\ln u+1\right)du \\\\&=\int_{0}^{\infty}e^{-u}\ln u\:du+1 \end{align} yielding $$\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}dx=-2\int_{0}^{\infty}e^{-u}\ln u\:du,$$ which is the announced result.
Note that $$\begin{eqnarray*} \frac{d}{du}\left[e^{-u}\left(u\ln u-(1+u)\ln(1+u)\right)\right]\hspace{-40mm}&&\\ &=&e^{-u}\left(\ln u-\ln(1+u)-u\ln u+(1+u)\ln(1+u)\right)\\ &=&e^{-u}\left(\ln u+u\ln(1+1/u)\right). \end{eqnarray*}$$ Hence $$\begin{eqnarray*} 2\int_0^\infty ue^{-u}\ln(1+1/u)\;du &=&2\left[e^{-u}\left(u\ln u-(1+u)\ln(1+u)\right)\right]_0^\infty\\ &&{}-2\int_0^\infty e^{-u}\ln u\;du\\ &=&2\gamma \end{eqnarray*}$$ by the first formula here.
Starting from $$2\int_0^\infty u e^{-u}\ln\left(1+\frac{1}{u}\right)du =2\int_0^\infty u e^{-u}\left(\ln(1+u)-\ln(u)\right)du$$ Let's find $\int_0^\infty u e^{-u}\ln(u)du$ first. We can introduce the variable $a$ and examine the integral: $$\int_0^\infty e^{-u} u^adu=\Gamma(a+1)$$ Now we will take the derivative by $a$ to get back to our original integral: $$\frac{\partial}{\partial a}\int_0^\infty e^{-u} u^adu =\int_0^\infty e^{-u}u^a\ln(u)du=\Gamma'(a+1)$$ Let $a=1$. We then have $$\int_0^\infty e^{-u}u\ln(u)du=\Gamma'(2)=1-\gamma$$ Now $$\int_0^\infty ue^{-u}\ln(1+u)du=\left[-e^{-u}u\ln(1+u)\right]_0^\infty+ \int_0^\infty e^{-u}\left(\ln(1+u)+\frac{u}{1+u}\right)du$$ by integration by parts. The edge terms are zero, so we are left with (integrating the first integrand by parts again) $$\int_0^\infty e^{-u}\ln(1+u)du+\int_0^\infty e^{-u}\frac{u}{1+u}du$$ $$=[-e^{-u}\ln(1+u)]_0^\infty+\int_0^\infty \frac{e^{-u}}{1+u}du+\int_0^\infty e^{-u}\frac{u}{1+u}du$$ $$=0+\int_0^\infty e^{-u}du = 1$$ Thus, combining this with our previous result in the first integral: $$2\int_0^\infty u e^{-u}\ln\left(1+\frac{1}{u}\right)du=2(1-(1-\gamma))=2\gamma$$
Let $I$ your integral (2) (without the factor $2$). As $1-(x+1)\exp(-x)$ is a primitive of $x\exp(-x)$, an integration by parts gives $$I=\int_0^{+\infty} (\frac {1}{1+x}-\exp(-x))\frac{dx}{x}$$ Then use http://mathworld.wolfram.com/Euler-MascheroniConstant.html