Why is $\int_0^1 \frac{t^x - 1}{\log(t)} dt$ finite for $x \in [0,1]$? I want to show that the following integral
$$\int_0^1 \frac{t^x - 1}{\log(t)} dt$$ 
is finite for all $x \in [0,1]$. But I struggle to come up with a good argument for that. Thanks in advance :)
 A: 
I thought it might be instructive to present a way forward that relies on only a set of elementary inequalities.  To that end, we proceed.


First, we can write the integrand as
$$\frac{t^x-1}{\log(t)}=\frac{e^{x\log(t)}-1}{\log(t)}$$
Next, we use the inequalities for the expoential function, $1+x\le e^x\le \frac1{1-x}$, for $x<1$, which I established in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's inequality.  This reveals that for $0<t< 1$, $0\le x\le 1$

$$x\ge \frac{e^{x\log(t)}-1}{\log(t)}\ge \frac{x}{1-x\log(t)}\ge 0$$

for $x\log(t)<1$, which is the case for $x\in [0,1]$ and $t\in (0,1)$.  
Hence, the integral, $\displaystyle \int_0^1 \frac{t^x-1}{\log(t)}\,dt$, not only converges, but is bounded below by $0$ and above by $x$.

Moreover, note that $\frac{t^x-1}{\log(t)}=\int_0^x t^y\,dy$ for $t>0$.  Therefore, we can write for $x\ge 0$
$$\begin{align}
\int_0^1 \frac{t^x-1}{\log(t)}\,dt&=\int_0^1\int_0^x t^y\,dy\,dt\\\\
&=\int_0^x\int_0^1t^y\,dt\,dy\\\\
&=\int_0^x \frac1{y+1}\,dy\\\\\
&=\log(x+1)
\end{align}$$
And certainly, $0\le \log(x+1)\le x$ for $x\ge 0$ as we ascertained.
A: With the substitution $t=e^{-z}$ we get
$$ \int_{\varepsilon}^{1}\frac{t^x-1}{\log t}\,dt=\int_{0}^{-\log\varepsilon}\frac{1-e^{-xz}}{z}e^{-z}\,dz $$
and through the dominated convergence theorem and Frullani's theorem we get that the limit as $\varepsilon\to 0^+$ is just $\color{red}{\log(x+1)}$ for any $x>-1$.
A: The function $t\mapsto \frac{t^x-1}{\ln t}$ has a finite limit at $0$ equal to $0$ so it's an extended function by continuity at $0$ and then it's integrable in a neighborhood of $0$. Moreover, by change of variable $u=1-t$ we get
$$\frac{(1-u)^x-1}{\ln(1-u)}\sim_0-\frac{x\ln(1-u)}{u}\sim x $$
so the function has also a finite limit at $1$. Conclude.
A: IF we differentiate under the integral sign we have
\begin{array}
$ I(x)&=&\displaystyle\int_0^1\frac{t^x-1}{\log t}dt\\
I'(x)&=&\displaystyle\int_0^1\frac{t^x\log t}{\log t}dt\\
&=&\displaystyle\int_0^1 t^x dt\\
&=&\displaystyle\frac{1}{1+x}, \quad x\neq -1
\end{array}
Next integrating we have $I(x)=\log(1+x)+c$. Since $I(0)=0$, $c=0$, we are done.
