Why is $\int_0^1 \frac{t^x - 1}{\log(t)} dt$ finite for $x \in [0,1]$?

Question

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 :)

Answer 1

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.

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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 $x$.

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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.

Answer 2

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 $x>-1$.

Answer 3

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.

Answer 4

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.