calculus - 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.

Answer

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.