Show that the function $ G:(0,1) \rightarrow \mathbb{R}, x \mapsto \int \limits_{0}^{1} \frac{t^{x}-1}{\log t} d t $ is continuously differentiable Show that the function $ G:(0,1) \rightarrow \mathbb{R}, x \mapsto \int \limits_{0}^{1} \frac{t^{x}-1}{\log t} d t $ is continuously differentiable and calculate the derivative explicitly.
Problem/approach:
I already know how to calculate the explicit derivative here, but when it comes to justifying why this parameter integral is continuously differentiable, I don't really know what to write there.
I know that there are two theorems for the parameter integral, where you can infer them from the integrand, but the problem is that these are defined on compact intervals.So how should you justify that then, if this condition is notbpresent? Here (0,1) is no compact interval!
Here is the calculation of the derivative with respect to x for review:
$\begin{aligned} \frac{\partial}{\partial x} \int \limits_{0}^{1} \frac{t^{x}-1}{\ln (t)} d t=\int \limits_{0}^{1} \frac{\partial}{\partial x} \frac{t^{x}-1}{\ln (t)} d t=\int \limits_{0}^{1} \frac{\ln (t) t^{x}}{\ln (t)} d t=\int \limits_{0}^{1} t^{x} d t=\frac{1}{x+1} .\end{aligned} $
 A: You have it, the derivative is:
$$ \begin{aligned} \frac{\partial}{\partial x} \int \limits_{0}^{1} \frac{t^{x}-1}{\ln (x)} d t=\frac{1}{x+1} .\end{aligned} $$
Which is discontinuous only on $x=-1$ and therefore the derivate is continuous in $(0,1)$. Given that $x=-1\notin (0,1)$.
This may look easy, but there is nothing wrong with it.
Even if you want to use the Theorem that requires a compact interval, you can take the closure of $(0,1)$ which is $[0,1]$. Given that the derivative is continuous on $[0,1]$, and that $(0,1)\in[0,1]$, it follows that the derivative is continuous in $(0,1)$.
A: I shall find ,by the first principle, the first derivative of the function $G$ defined by
$$
G(x)=\int_{0}^{1} \frac{t^{x}-1}{\ln t} d t
$$
$$
\begin{aligned}
\lim _{h \rightarrow 0} \frac{G(x+h)-G(x)}{h} 
=& \lim _{h \rightarrow 0} \int_{0}^{1} \frac{1}{h}\left(\frac{t^{x+h}-1}{\ln t}-\frac{t^{x}-1}{\ln t}\right) d t \\
=& \lim _{h \rightarrow 0} \int_{0}^{1} \frac{1}{h} \frac{t^{x}\left(t^{h}-1\right)}{\ln t} d t
\\=&\int_{0}^{1} \frac{t^{x}}{\ln t}\left(\lim _{h \rightarrow 0} \frac{t^{h}-1}{h}\right) d t 
\\=&\int_{0}^{1} \frac{t^{x}}{\ln t} \ln t d t \\
=&\frac{1}{x+1} \quad \text { if } x \neq-1 
\end{aligned}
$$
Hence G is differentiable for any $x\in (0,1)$. Furthermore, G is continuously differentiable on (0,1) and $$
G^{(n)}(x)=\frac{(-1)^{n+1} n !}{(x+1)^{n+1}}
$$
