Finding the integral $\int_0^1 \frac{x^a - 1}{\log x} dx$
How to do the following integral:
$$\int_{0}^1 \dfrac{x^a-1}{\log(x)}dx$$ where $a \geq 0$?
I was asked this question by a friend, and couldn't think of any substitution that works. Plugging in a=2,3, etc in Wolfram, I get values like $\log(a+1)$, which may be the right answer (for general $a$). Is there a simple way to calculate this integral?
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$\begingroup$Call your integral $I(a)$. Then $$ I'(a) = \int_0^1 x^a dx = \frac{1}{a+1} $$ as long as $a \geq 0$. Now you need to solve the differential equation $$ I'(a) = \frac{1}{a + 1}.$$ This is a very easy differential equation to solve, and the solution is $$ I(a) = \log(a+1) + C $$ where $C$ is some constant. Now we ask, what is that constant? Notice that $$ I(0) = \int_0^1 \frac{1 - 1}{\log x} dx = 0,$$ so we need $$ I(0) = \log(1) + C = 0,$$ or rather $C = 0$. So we conclude that $$ \int_0^1 \frac{x^a - 1}{\log x} dx = \log(a + 1), $$ as you suggested. $\diamondsuit$
$\endgroup$ 2 $\begingroup$We can utilize $$ \int_0^1x^t\,\mathrm{d}t=\frac{x-1}{\log(x)} $$ combined with the substitution $x\mapsto x^{1/a}$, to get $$ \begin{align} \int_0^1\frac{x^a-1}{\log(x)}\,\mathrm{d}x &=\int_0^1\frac{x-1}{\log(x)}x^{\frac1a-1}\,\mathrm{d}x\\ &=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_0^1\frac1{\frac1a+t}\,\mathrm{d}t\\ &=\log\left(\frac1a+1\right)-\log\left(\frac1a\right)\\[9pt] &=\log(1+a) \end{align} $$
$\endgroup$ $\begingroup$A couple of possibilities:
Change variables to $x=e^{-y}$, $dx = -e^{-y} \, dy$, so the integral becomes $$ \int_0^{\infty} \frac{e^{-y}-e^{-(1+a)y}}{y} \, dy. $$ This is a Frullani integral, so the value is $$ (1-0)\log{(1+a)/1} = \log{(1+a)}. $$
Differentiate under the integral sign. I won't go through this one since someone else got there first.
Do the substitution from 1., but integrate by parts: $$ I(a) = \left[ (e^{-y}-e^{-(1+a)y})\log{y} \right] + \int_0^{\infty} e^{-y}\log{y} \, dy - (1+a)\int_0^{\infty} e^{-(1+a)y} \log{y} \, dy, $$ The first term is zero since the bracket is $O(y)$ at $0$. Change variables to $u=(1+a)y$ in the last integral. It turns into $$ -\int_0^{\infty} e^{-u} (\log{u}-\log{(1+a)} \, dy = -\int_0^{\infty} e^{-y}\log{y} \, dy + \log{(1+a)} $$, and then the remaining integrals cancel and you get the result.
We have $x^a-1 = e^{a\log(x)}-1$. Hence, the integral is \begin{align} I & = \int_0^1 \dfrac{x^a-1}{\log(x)}dx = \int_0^1 \left(\sum_{k=1}^{\infty} \dfrac{a^k \log^k(x)}{k!}\right)\dfrac{dx}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} \int_0^1 \log^{k-1}(x)dx\\ & = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} (-1)^{k-1} (k-1)! = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}a^k}k = \log(1+a) \end{align}
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