$\Gamma\in C^\infty(0,\infty)$. How does one demonstrate that the Euler $\Gamma$-function $\Gamma:(0,\infty)\rightarrow\mathbf{R}$ is of class $C^\infty$, taking the integral to be an improper Riemann-integral? I only know how to differentiate proper Riemann integrals (differentiation under the integral sign). No complex analysis or measure theory arguments are allowed.
 A: Here is a sketch of proof that meets (I hope) the requirements. I concentrate on the proof that $\Gamma$ is $\mathcal C^1$; the reasoning is the same for higher derivatives.
First, let us recall the definition:
$$\Gamma(x)=\int_0^\infty t^{x-1}\, e^{-t}dt\, . $$
This can be written as 
$$\Gamma(x)=\lim_{n\to\infty}\int_{\frac1n}^nt^{x-1}\, e^{-t}dt:=\lim_{n\to\infty} \Gamma_n(x)\, . $$
By the "standard" theorem on differentiation under the integral sign for proper Riemann integrals, the functions $\Gamma_n$ are $\mathcal C^1$ with 
$$\Gamma'_n(x)=\int_{\frac1n}^n\log(t)\, t^{x-1}\, e^{-t}dt\, . $$
To prove that $\Gamma=\lim \Gamma_n$ is $\mathcal C^1$, it is enough to show that the sequence of derivatives $(\Gamma_n'(x))$ is pointwise convergent to some function $g:(0,\infty)\to\mathbb R$, and that the convergence is uniform over any compact interval $[a,b]\subset (0,\infty)$.
The first point is clear because the improper Riemann integral $$g(x)=\int_0^\infty \log(t)\, t^{x-1}e^{-t}dt$$ is easily seen to be convergent, and in fact absolutely convergent. What has to be done is to check the uniform convergence.
So let us fix a compact interval $[a,b]\subset (0,\infty)$. For any $n\in\mathbb N$ and $x\in[a,b]$ we have
\begin{eqnarray}\Gamma'_n(x)-g(x)&=&\int_0^{\frac1n} \log(t)\, t^{x-1}e^{-t}dt+\int_n^\infty \log(t)\, t^{x-1}e^{-t}dt\\
&=&\lim_{\varepsilon\to 0^+}\int_\varepsilon^{\frac1n} \log(t)\, t^{x-1}e^{-t}dt+\lim_{X\to\infty}\int_n^X \log(t)\, t^{x-1}e^{-t}dt\, .
\end{eqnarray}
Now, for $0<\varepsilon< \frac1n$ and $X>n$ we also have
$$\left\vert\int_\varepsilon^{\frac1n} \log(t)\, t^{x-1}e^{-t}dt\right\vert\leq \int_0^{\frac1n} \vert\log(t)\vert\, t^{x-1}e^{-t}dt\, ,$$
which makes sense because the integral $g(x)$ is absolutely convergent. Since $t\leq 1$ for every $t\in(0,\frac1n]$ and since $x\geq a$, it follows that
$$\left\vert\int_\varepsilon^{\frac1n} \log(t)\, t^{x-1}e^{-t}dt\right\vert\leq \int_0^{\frac1n} \vert\log(t)\vert\, t^{a-1}e^{-t}dt\, .$$
In the same way, we get 
\begin{eqnarray}\left\vert\int_n^X \log(t)\, t^{x-1}e^{-t}dt\right\vert&\leq& \int_n^\infty \vert\log(t)\vert\, t^{x-1}e^{-t}dt\\
&\leq& \int_n^\infty \vert\log(t)\vert\, t^{b-1}e^{-t}dt\, .
\end{eqnarray}
Altogether, this gives 
$$\vert \Gamma'_n(x)-g(x)\vert\leq \int_0^{\frac1n} \vert\log(t)\vert\, t^{a-1}e^{-t}dt+\int_n^\infty \vert\log(t)\vert\, t^{b-1}e^{-t}dt\, .  $$
The right-hand side does not depend on $x\in[a,b]$, and it tends to $0$ as $n\to\infty$ because the integral $g(x)$ is absolutely convergent. This proves the uniform convergence of $\Gamma'_n(x)$ to $g(x)$ on any compact interval $[a,b]\subset (0,\infty)$, and hence that $\Gamma$ is $\mathcal C^1$. 
