Computing $\int_{0}^{1} \frac{x^{t}-1}{\ln x} d x$ without Feynman, where $Re(t)>-1$?

Question

Differentiating the integral w.r.t. $a$ yields $$ \begin{aligned} \frac{\partial}{\partial a} \int_{0}^{1} \frac{x^{a}-1}{\ln x} d x =\int_{0}^{1} \frac{x^{a} \ln x}{\ln x} d x =\left[\frac{x^{a+1}}{a+1}\right]_{0}^{1} &=\frac{1}{a+1} \end{aligned} $$

Integrating both sides from $a=0$ to $a=t$ yields the result

$$ \boxed{\int_{0}^{1} \frac{x^{t}-1}{\ln x}dx=\ln (t+1)} $$ where $Re(t)>-1.$

Question: Can it be evaluated without Feynman?

Note${\int_{0}^{1} \frac{x^{t}-1}{\ln x}dx=\ln |t+1|}$ is false for $t\le-1$. — GEdgar, Jun 25 '22 at 02:23
After $u=-\ln x$ it's a Frullani integral $\int_0^\infty\frac{f(au)-f(bu)}{u}du$ with $a:=t+1,,b:=1,,f(z):=1-e^{-z}$. — J.G., Jun 25 '22 at 11:37
And if $t>-1$ you may omit the absolute value. Probably you get a nice result if you allow $t$ complex and require $\operatorname{Re}t>-1$. Of course then you must omit the absolute value. — GEdgar, Jun 25 '22 at 13:32

score 7 · Accepted Answer · edited Jun 26 '22 at 03:13

7

Sure, we can use the dual to Feynman - turning the integral into a double integral:

$$\int_0^1\frac{x^t-1}{\log x}\:dx = \int_0^1\int_0^tx^y\:dy\:dx = \int_0^t \frac{1}{y+1}\:dy = \log(t+1)$$

by Fubini's theorem.

edited Jun 26 '22 at 03:13

Lai

20,421

answered Jun 25 '22 at 02:29

Ninad Munshi

34,407

score 1 · Answer 2 · answered Jun 25 '22 at 06:03

Using special functions $$I=\int\frac{x^{t}-1}{\log( x)}\,d x=\int\frac{e^{(t+1) y}}{y}\,dy-\int\frac{e^y}{y}\,dy=\text{Ei}((t+1) y)-\text{Ei}(y)=\text{Ei}((t+1) \log (x))-\text{li}(x)$$

$$\lim_{x\to 0} \, \Big[\text{Ei}((t+1) \log (x))-\text{li}(x)\Big] =0$$

$$J_k=\int_0^{e^k}\frac{x^{t}-1}{\log( x)}\,d x=\text{Ei}(k(t+1))-\text{Ei}(k)$$ The only simple case is for $k=0$ $$J_0=\frac{1}{2} \left(\log (t+1)-\log \left(\frac{1}{t+1}\right)\right)=\log(t+1)$$

score 1 · Answer 3 · answered Jun 25 '22 at 11:58

This is an answer from Hints on calculating the integral $\int_0^1\frac{x^{19}-1}{\ln x}\,dx$ of mine for $t=m$ an integer. Note $$ \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln x, \text{ for }x>0 $$ and hence, for $m\in\mathbb{N}$, \begin{eqnarray} \int_0^1\frac{x^m-1}{\ln x}dx&=&\int_0^1\sum_{i=0}^{m-1}\frac{x^i(x-1)}{\ln x}dx\\ &=&\int_0^1\sum_{i=0}^{m-1}\lim_{n\to\infty}\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\ &=&\lim_{n\to\infty}\sum_{i=0}^{m-1}\int_0^1\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\ &=&\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{m-1}\int_0^1x^i\sum_{k=0}^{n-1}x^{\frac{k}{n}}dx\\ &=&\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{m-1}\int_0^1\sum_{k=0}^{n-1}x^{\frac{k}{n}+i}dx\\ &=&\lim_{n\to\infty}\sum_{i=0}^{m-1}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{k}{n}+i+1}\\ &=&\sum_{i=0}^{m-1}\int_0^1\frac{1}{x+i+1}dx\\ &=&\sum_{i=0}^{m-1}\ln\frac{i+2}{i+1}\\ &=&\ln(m+1). \end{eqnarray}

score 0 · Answer 4 · answered Jul 03 '22 at 03:41

$$ \begin{aligned} \int_{0}^{1} \frac{x^{t}-1}{\ln x} d x &\stackrel{x \mapsto e^{-x}}{=} \int_{0}^{\infty} \frac{e^{-x t} -1}{x} e^{-x} d x \\ &=\int_{0}^{\infty} \frac{e^{-x(t+1)}-e^{-x}}{x} d x \end{aligned} $$

Using the Frullani’s Theorem, we have $$ \int_{0}^{\infty} \frac{e^{-x(t+1)}-e^{-x}}{x} d x=(0-1) \ln \left(\frac{-1}{-(t+1)}\right)=\ln (t+1) $$

Computing $\int_{0}^{1} \frac{x^{t}-1}{\ln x} d x$ without Feynman, where $Re(t)>-1$?

4 Answers4