Showing that $ \int_0^1 \frac{t^{x}-1}{\ln(t)}dt$ exists for all $x$ How do I show that $\displaystyle  \int_0^1 \frac{t^{x}-1}{\ln(t)}\,dt $ exists for every $ x \in \mathbb{R} $?
I have tried bounding it but without success...
Notice that the function $f:t\mapsto \frac{t^x-1}{\ln t}$ can be extended by continuity on 1 since by the Hospital rule
$$
\lim_{t\to1}f(t)=\lim_{t\to1}\frac{xt^{x-1}}{1/t}=x,
$$
hence the integral $\int_{1/2}^1f(t)dt$ exists for all $x$. Now for the problem in 0 and by change of variable $u=-\ln t$ the problem becomes at $\infty$
$$\int_0^1 f(t)dt=\int_{0}^\infty\frac{e^{-u}-e^{-(x+1)u}}{u}du$$
so because the integral $\int_1^\infty \frac{e^{-u}}{u}du$ is convergent we see that the given integral converges iff $x+1&gt;0$ i.e. $x&gt;-1$.
Define $$f(x)=\int_0^1\frac{t^x-1}{\log t}\,dt.$$
Then, we have
\begin{align}
f'(x)&amp;=\int_0^1\frac{\partial}{\partial x}\frac{t^x-1}{\log t}\,dt\\
f'(x)&amp;=\int_0^1t^x\,dt\\
f'(x)&amp;=\frac{1}{x+1}, \quad x&gt;-1.\\
\end{align}
Therefore, the statement is wrong.