Prove the uniform convergence involving logarithm The function is like:
$$F(x)=\int_{0}^{+\infty}(\ln t) (t^{x-1} )(e^{-t}) dt $$
Prove the integration's uniform convergence on $(0,+\infty)$

Well, you may notice that this is the derivative of $\Gamma$ function. The part I feel hard is how to process the $\ln t$ in it? Whether or not could you give the whole process, Thank you a lot!

 A: The integral $F(x)$ converges uniformly for $x \in [a,b]$ with $b> a >0$.
Consider, separately, the integrals over intervals $[1,\infty)$ and $[0,1]$.
Note that for $x \in [a,b]$ we have $|\ln(t)t^{x-1}e^{-t}| \leq \ln(t)t^{b-1}e^{-t}$ and 
$$\lim_{t \rightarrow \infty}\frac{(\ln t)t^{b-1}e^{-t}}{t^{-2}}= 0,$$
Hence, there exists $c>0$ and $K>0$ such that $(\ln t)t^{b-1}e^{-t}< ct^{-2}$ for $x \in [a,b] , and $
$$\int_{K}^{\infty} (\ln t)t^{x-1}e^{-t}dt < \int_{K}^{\infty} t^{-2}dt = \frac1{K}.$$
It follows by the $M$-test that the integral over $[1,\infty)$ converges uniformly for $x \in [a,b]$. 
For $t \in (0,1]$ and $x \geq a > 0$,
$$|(\ln t)t^{x-1}e^{-t}|< -(\ln t)t^{a-1}.$$
Hence, 
$$\int_{0}^{1} |(\ln t)t^{x-1}e^{-t}|dt < -\int_{0}^{1} (\ln t)t^{a-1}dt = \frac1{a^2}$$
and the integral over $[0,1]$ converges uniformly for $x \in [a,\infty).$
However, the integral $F(x)$ does not converge uniformly on the 0pen interval $(0,\infty).$  We have for any $\delta$ with $0 < \delta < 1$,
$$\left|\int_{0}^{\delta} (\ln t)t^{x-1}e^{-t}dt\right| > e^{-1}\int_{0}^{\delta} (-\ln t)t^{t-1}dt =  e^{-1}\frac{\delta^x (1-x\ln \delta)}{x^2}.$$
Since, for fixed $\delta$
$$\lim_{x \rightarrow 0}\frac{\delta^x (1-x\ln \delta)}{x^2} = \infty$$
we can find $x_\delta$ such that
$$\left|\int_{0}^{\delta} (\ln t)t^{x_\delta-1}e^{-t}dt\right| > 1.$$
