# Thread: Tough Integral

1. ## Tough Integral

For $\displaystyle m\in\mathbb{N}$, find the following integral:

$\displaystyle I_{m} = \int_{0}^{1}\left(\frac{\ln{x}}{1-x}\right)^{m}\;{dx}.$

Source: a cool guy called Drexel28 left it as an exercise for an unsuspecting student in mathlinks.

2. ## Re: Tough Integral

Originally Posted by TheCoffeeMachine
For $\displaystyle m\in\mathbb{N}$, find the following integral:

$\displaystyle I_{m} = \int_{0}^{1}\left(\frac{\ln{x}}{1-x}\right)^{m}\;{dx}.$

Source: a cool guy called Drexel28 left it as an exercise for an unsuspecting student in mathlinks.
Several years ago I 'discovered' that is...

$\displaystyle \int_{0}^{1} x^{n}\ \ln ^{m} x\ dx = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}}$ (1)

Now if we consider the 'binomial expansion'...

$\displaystyle (1-x)^{-m} = 1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!}\ x^{n}$ (2)

... combining (1) and (2) we obtain...

$\displaystyle I_{m}= \int_{0}^{1} (\frac{\ln x}{1-x})^{m}\ dx = (-1)^{m}\ m!\ \{1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!\ (n+1)^{m+1}} \}$ (3)

At this point some more work on the series (3) is useful!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Tough Integral

Originally Posted by chisigma
Several years ago I 'discovered' that is...

$\displaystyle \int_{0}^{1} x^{n}\ \ln ^{m} x\ dx = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}}$ (1)

Now if we consider the 'binomial expansion'...

$\displaystyle (1-x)^{-m} = 1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!}\ x^{n}$ (2)

... combining (1) and (2) we obtain...

$\displaystyle I_{m}= \int_{0}^{1} (\frac{\ln x}{1-x})^{m}\ dx = (-1)^{m}\ m!\ \{1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!\ (n+1)^{m+1}} \}$ (3)

At this point some more work on the series (3) is useful!...
For some values of m...

$\displaystyle I_{1} = - \{1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}}\} = - \zeta(2)$ (1)

$\displaystyle I_{2} = 2 \{1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}}\} = 2\ \zeta(2)$ (2)

$\displaystyle I_{3} = - 3! \{1 + \sum_{n=1}^{\infty} \frac{n+2}{2\ (n+1)^{3}}\} = - 3! \{1 + \sum_{n=1}^{\infty} \frac{1}{2\ (n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{2\ (n+1)^{3}} \} =$

$\displaystyle = -3! - 3\ \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}} - 3\ \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} = -3\ \{\zeta(2) + \zeta(3)\}$ (3)

Does anyone want to proceed?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. ## Re: Tough Integral

Originally Posted by chisigma
Several years ago I 'discovered' that is...

$\displaystyle \int_{0}^{1} x^{n}\ \ln ^{m} x\ dx = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}}$ (1)
Nice. I've seen you use this before. Did you prove by setting up a recurrence relation?

5. ## Re: Tough Integral

Gamma function,

$\displaystyle \int_{0}^{1}{{{x}^{n}}{{\ln }^{m}}x\,dx}={{(-1)}^{m}}\int_{0}^{\infty }{{{t}^{m}}{{e}^{-(n+1)t}}\,dt}=\frac{{{(-1)}^{m}}}{{{(n+1)}^{m+1}}}\int_{0}^{\infty }{{{t}^{m}}{{e}^{-t}}\,dt}=\frac{{{(-1)}^{m}}m!}{{{(n+1)}^{m+1}}}.$