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.