I'm trying to prove that the function
$\varphi(x)= \int_0^\infty \frac{ t^{x-1} }{1+t} dt$
is continuous for any $0<x<1$.
I guess I have to use LDCT, and show something like
$\int_0^\infty \frac{1}{t^{\alpha}(1+t)} dt < \infty$,
for any $\alpha \in (0,1)$.
Any ideas on how to solve the question, or at least on how to show that the later integral is indeed convergent.
$\phi (x)$ does not exist for any $x$.
Perhaps, the denominator is supposed to be $1+t$. In that case we can prove continuity of $\phi$ as follows:
Let $x_n \to x \in (0,1)$. Then, for $t \geq 1$ and $n$ sufficiently large we have $|\frac {t^{x_n-1}} {1+t}|\leq \frac {t^{(x-1)}/2} {t+1}$ which is integrable in $(1,\infty)$. For $t <1$ and $n$ sufficiently large we have $|\frac {t^{x_n-1}} {1+t}|\leq \frac {t^{x/2-1}} {t+1}$ which is integrable on $(0,1)$. Now apply DCT to each of the intervals $(0,1)$ and $(1,\infty)$.