Proving $ \int_{0}^{\infty} \frac{\ln(t)}{\sqrt{t}}e^{-t} \mathrm dt=-\sqrt{\pi}(\gamma+\ln{4})$

I would like to prove that:

$$ \int_{0}^{\infty} \frac{\ln(t)}{\sqrt{t}}e^{-t} \mathrm dt=-\sqrt{\pi}(\gamma+\ln{4})$$

I tried to use the integral $$\int_{0}^{n} \frac{\ln(t)}{\sqrt{t}}\left(1-\frac{t}{n}\right)^n \mathrm dt$$

$$\int_{0}^{n} \frac{\ln(t)}{\sqrt{t}}\left(1-\frac{t}{n}\right)^n \mathrm dt \;{\underset{\small n\to\infty}{\longrightarrow}}\; \int_{0}^{\infty} \frac{\ln(t)}{\sqrt{t}}e^{-t} \mathrm dt$$ (dominated convergence theorem)

Using the substitution $t\to\frac{t}{n}$, I get:

$$ \int_{0}^{n} \frac{\ln(t)}{\sqrt{t}}\left(1-\frac{t}{n}\right)^n \mathrm dt=\sqrt{n}\left(\ln(n)\int_{0}^{1} \frac{(1-t)^n}{\sqrt{t}} \mathrm dt+\int_{0}^{1} \frac{\ln(t)(1-t)^n}{\sqrt{t}} \mathrm dt\right) $$

However I don't know if I am on the right track for these new integrals look quite tricky.

Solution 1:

Consider integral representation for the Euler $\Gamma$-function: $$ \Gamma(s) = \int_0^\infty t^{s-1} \mathrm{e}^{-t} \mathrm{d} t $$ Differentiate with respect to $s$: $$ \Gamma(s) \psi(s) = \int_0^\infty t^{s-1} \ln(t) \mathrm{e}^{-t} \mathrm{d} t $$ where $\psi(s)$ is the digamma function. Now substitute $s=\frac{1}{2}$. So $$ \int_0^\infty \frac{ \ln(t)}{\sqrt{t}} \mathrm{e}^{-t} \mathrm{d} t = \Gamma\left( \frac{1}{2} \right) \psi\left( \frac{1}{2} \right) $$ Now use duplication formula: $$ \Gamma(2s) = \Gamma(s) \Gamma(s+1/2) \frac{2^{2s-1}}{\sqrt{\pi}} $$ Differentiating this with respect to $s$ gives the duplication formula for $\psi(s)$, and substitution of $s=1/2$ gives $\Gamma(1/2) = \sqrt{\pi}$. $$ \psi(2s) = \frac{1}{2}\psi(s) + \frac{1}{2} \psi(s+1/2) + \log(2) $$ Substitute $s=\frac{1}{2}$ and use $\psi(1) = -\gamma$ to arrive at the result.

Solution 2:

I don't have enough time to write a full solution (I will when I find the time), but hopefully this skeletal answer is sufficiently illustrative.

The idea is to use the Frullani integral

$$\log\frac{b}{a}=\int_0^\infty \frac{\exp(-at)-\exp(-bt)}{t}\mathrm dt$$

and substitute this into the original integral:

$$\int_0^{\infty} \frac{\exp(-t)}{\sqrt{t}} \log\,t\;\mathrm dt=\int_0^{\infty} \frac{\exp(-t)}{\sqrt{t}} \left(\int_0^\infty \frac{\exp(-u)-\exp(-tu)}{u}\mathrm du\right)\mathrm dt$$

or

$$\int_0^{\infty}\frac1{u}\int_0^\infty \frac{\exp(-t-u)-\exp(-t(1+u))}{\sqrt{t}}\mathrm dt\mathrm du$$

which expands to

$$\int_0^{\infty}\frac1{u}\left(\exp(-u)\int_0^\infty \frac{\exp(-t)}{\sqrt{t}}\mathrm dt-\int_0^\infty \frac{\exp(-t(1+u))}{\sqrt{t}}\mathrm dt\right)\mathrm du$$

There is the Laplace transform relation

$$\int_0^\infty \frac{\exp(-cv)}{\sqrt v}\mathrm dv=\sqrt{\frac{\pi}{c}}$$

which we can use here to yield

$$\sqrt{\pi}\int_0^{\infty}\frac1{u}\left(\exp(-u)-\frac1{\sqrt{1+u}}\right)\mathrm du$$

Integration by parts yields the equivalent

$$-\sqrt{\pi}\left(\frac12\int_0^{\infty}\frac{\log\,u}{(1+u)^\frac32}\mathrm du\color{blue}{-\int_0^{\infty}\exp(-u)\log\,u\mathrm du}\right)$$

and the portion highlighted in blue is recognized to be the Euler-Mascheroni constant (see this as well):

$$-\sqrt{\pi}\left(\frac12\int_0^{\infty}\frac{\log\,u}{(1+u)^\frac32}\mathrm du+\color{blue}{\gamma}\right)$$

Still another application of integration by parts yields

$$-\sqrt{\pi}\left(\frac12\left.{\left(\frac{2\log\,u}{\sqrt{u+1}}-2\log \left(1+\frac{2\left(1-\sqrt{1+u}\right)}{u}\right)\right)}\right|_0^\infty+\gamma\right)$$

and then one can use e.g. l'Hôpital's rule to evaluate the limit implicit in the expression above (for $u\to 0$ that is; for $u\to\infty$, no special treatment is needed and the logarithmic expression evaluates to zero) to finally yield the expected $-\sqrt{\pi}(\log\,4+\gamma)$.