5

$$\int_{0}^{1}{x^n-1 \over \ln(x)}dx=\ln(n+1)$$

Let deal with case $n=1$

$$I=\int_{0}^{1}{x-1 \over \ln(x)}dx=\ln(2)$$

$u=\ln(x)$ $\rightarrow du=\frac{1}{x}dx$

$x \rightarrow 1 ,u=0$

$x \rightarrow 0, u=-\infty$

$$I=-\int_{0}^{\infty}\frac{e^{2u}-e^u}{u}du$$

Apply integration by parts

$$I=\left.(e^{2u}-e^u)\ln(u)\right|_{0}^{\infty}-\int_{0}^{\infty}(2e^{2u}-e^u)\ln(u)du$$

Letting

$$J=\int_{0}^{\infty}(2e^{2u}-e^u)\ln(u)du$$

Applying by parts again

$$J=\left.(2e^{2u}-e^u)\ln(u)\right|_{0}^{\infty}-\int_{0}^{\infty}(2e^{2u}-e^u)\frac{1}{u}du$$

Anyway I skip the simplification and get to the result

$$2I=\left.e^{2u}\ln(u)\right|_{0}^{\infty}+\int_{0}^{\infty}\frac{e^{2u}}{u}du$$

That doesn't looked correct!

Integration by parts and substitution seem to failed here for me, so what is another method to evaluate this integral?

Another attempt using $x^n-1=\sum_{k=0}^{n-1}(x-1)x^k$

$$\int_{0}^{1}{x^n-1 \over \ln(x)}dx=\ln(n+1)$$

$$\sum_{k=0}^{n-1}\int_{0}^{1}{(x-1)x^k \over \ln(x)}dx=\ln(n+1)$$

This is still involving Integration by parts, I am very sure is going to lengthy so I am stopping here for help. Please lend me a hand, thank you.

Here is a link to Frullani's formula

gymbvghjkgkjkhgfkl
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  • This can't be duplicate, because here I asked specific on the general of the formula. I am not asking for specific $x^7$ – gymbvghjkgkjkhgfkl May 31 '16 at 16:47
  • If you actually read the answers posted in that question, you will find the general solution. – heropup May 31 '16 at 16:48
  • I haven't read so, ok it is duplicate thank you. – gymbvghjkgkjkhgfkl May 31 '16 at 16:50
  • Further duplicates to prove my point: http://math.stackexchange.com/questions/1263568/how-to-calculate-the-integral-i-int-limits-01-fracxn-1-lnx-mathrm-d?lq=1 http://math.stackexchange.com/questions/1254548/an-integration-question-to-be-solved-without-using-differentiation-under-the-int http://math.stackexchange.com/questions/778278/hints-on-calculating-the-integral-int-01-fracx19-1-ln-x-dx?lq=1 – heropup May 31 '16 at 16:50

2 Answers2

10

Your conjecture is a trivial consequence of Frullani's theorem after the substitution $x=e^{-z}$.

Jack D'Aurizio
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7

An easy way to prove this (while also generalising your statement to any real number) is by using the technique of differentiating under the integral sign. Let $$f(\alpha)=\int_0^1 \frac{x^\alpha -1}{\log x} dx.$$ By differentiation under the integral we see that $$f'(\alpha)=\int_0^1 \frac{\partial}{\partial \alpha}\left(\frac{x^\alpha -1}{\log x}\right)dx=\int_0^1 x^\alpha dx=\frac{1}{1+\alpha}$$ hence $$f(\alpha)=\int \frac{d\alpha}{1+\alpha}=\log(1+\alpha)+C.$$ But we notice from the definition that $f(0)=0$ so it follows that $C=0$ which completes the proof.

user1892304
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    Feynmann comes to the rescue. +1 – Mark Viola May 31 '16 at 16:29
  • @user1892304 can you show me how you get from $\frac{\partial}{\partial \alpha}{x^{\alpha}-1 \over \log(x)}$ to $x^{\alpha}$ – gymbvghjkgkjkhgfkl May 31 '16 at 17:14
  • I may need to know to solve another problem similar to this one – gymbvghjkgkjkhgfkl May 31 '16 at 17:17
  • @Chinacat In this step we are taking the partial derivative with respect to $\alpha$ so we consider $x$ as a constant. Here $-1/ \log x$ is a constant so it goes away, and the derivative of $x^\alpha$ w.r.t. $\alpha$ is $x^\alpha \log x$ so the $\log x$'s cancel out leaving just $x^\alpha$. – user1892304 May 31 '16 at 17:47