# Hints on calculating the integral $\int_0^1\frac{x^{19}-1}{\ln x}\,dx$

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

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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

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.

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@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.

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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}

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

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