Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would be happy to get some hints on the following integral: $$ \int_0^1\frac{x^{19}-1}{\ln x}\,dx $$

share|improve this question
Thanks egreg for editing. I tried but did not find any online MathJax editor while there are many LaTex ones. Hope this site be equipped with this functionality soon. –  Xaqron May 2 at 12:34
This is really similar to this: math.stackexchange.com/questions/566475 –  JimmyK4542 Jun 18 at 18:19
In connection with the "Feynman method" for evaluating this, my answer to What are some good low-prerequisite examples for the heuristic advice “If you cannot prove it, prove something stronger.”? may be of interest. –  Dave L. Renfro Jun 18 at 18:27

4 Answers 4

up vote 10 down vote accepted

Let $x = e^{-y}$, we have

$$\int_0^1 \frac{x^{19} - 1}{\log x} dx = \int_0^\infty \frac{e^{-\color{blue}{1}y} - e^{-\color{orange}{20}y}}{y} dy$$ This is in the form of a Frullani's integral and one can read off the value of the integral as

$$( \color{red}{1} - \color{green}{0} )\log\left(\frac{\color{orange}{20}}{\color{blue}{1}}\right) = \log 20 \quad\text{ since }\quad e^{-y} = \begin{cases}\color{red}{1}, &y = 0\\ \color{green}{0}, & y \to \infty\end{cases}$$

If you really need to perform the integral yourself without using Frullani's integral directly, I'll recommend you look at answers of this question and learn the various proof there. A good exercise is translate the proof there to your particular case. This will get you familiar with the steps that need to evaluate this sort of integral.

share|improve this answer
@J.J. thanks, fixed. –  achille hui May 2 at 13:08

Differentiation of the integrand $$f(x,a) = \frac{x^a-1}{\log x}$$ with respect to $a$ gives $$\frac{\partial f}{\partial a} = x^a.$$ Therefore, $$I(a) = \int_{x=0}^1 f(x,a) \, dx$$ implies $$\frac{d I}{d a} = \int_{x=0}^1 x^a \, dx = \frac{1}{a+1}, \quad a > -1.$$ Integrating with respect to $a$ then yields $$I(a) = \log(a+1), \quad a > -1.$$ There are some omitted details, but this is an outline of the general solution.

share|improve this answer

Here is the details from @heropup's answer.

Let us generalize the problem. We will evaluate $$ \mathcal{I}(\alpha)=\int_0^1\frac{x^\alpha-1}{\ln x}\ dx\qquad;\qquad \alpha>-1.\tag1 $$ Now we apply Feynman's method (differentiate under the integral sign). Diferentiating both sides of $(1)$ yields \begin{align} \frac{\partial\mathcal{I}}{\partial\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{x^\alpha-1}{\ln x}\right]\ dx\\ \mathcal{I}'(\alpha)&=\int_0^1 x^\alpha\ dx\\ &=\left.\frac{x^{\alpha+1}}{\alpha+1}\right|_{x=0}^1\\ &=\frac{1}{\alpha+1}.\tag2 \end{align} Integrating $(2)$ yields \begin{align} \mathcal{I}(\alpha)&=\int\frac{1}{\alpha+1}\ d\alpha\\ &=\ln(\alpha+1)+C.\tag3 \end{align} In order to find out our constant of integration, we let $\alpha = 0$ so that our integrand is $0$, implying that $C = 0$. Letting $\alpha = 19$ will of course solve our original problem: \begin{align} \color{purple}{\int_0^1\frac{x^\alpha-1}{\ln x}\ dx}&\color{purple}{=\ln(\alpha+1)}\\ \int_0^1\frac{x^{19}-1}{\ln x}\ dx&=\large\color{blue}{\ln20}. \end{align}

share|improve this answer

Hint: Substitute $\ln x=t$. Use integration by parts formula.

share|improve this answer
I took it as z and unfortunately no progress. Does taking it as t makes it solvable? –  Xaqron May 2 at 12:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.