# Math Help - integral

1. ## integral

How do I find integrals like this?

$\int_{0}^{1}\frac{\log(x)}{x-1}dx$

2. ## Re: integral

If you mean what is the definition of this kind of improper integral: $f(x)=\frac{\log x}{x-1}$ is continuous in $(0,1]$ , as a consequence integrable in $[\epsilon,1]\;(0<\epsilon\leq 1)$ ) . Besides, $\lim_{x\to 0^+}f(x)=+\infty$ so $\int_0^1f(x)=\lim_{\epsilon\to 0^+}\int_{\epsilon}^1f(x)\;dx$ (by definition).

3. ## Re: integral

Originally Posted by AgentSmith
How do I find integrals like this?

$\int_{0}^{1}\frac{\log(x)}{x-1}dx$
Are you asked to solve this integral?

4. ## Re: integral

Originally Posted by sbhatnagar
Are you asked to solve this integral?
yes

5. ## Re: integral

Some time ago I 'discovered' the general formula...

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

... and if we take into account that...

$\frac{1}{1-x}= \sum_{m=0}^{\infty} x^{m}$ (2)

... we obtain...

$\int_{0}^{1} \frac{\ln x}{x-1}\ dx = \sum_{m=0}^{\infty} \frac{1}{(m+1)^{2}}= \frac{\pi^{2}}{6}$

Kind regards

$\chi$ $\sigma$

6. ## Re: integral

Here's what I did...

1.
\begin{align*} I=\int_{0}^{1}\frac{\ln(x)}{(x-1)}dx &= \int_{0}^{1} \frac{(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\cdots}{x-1}dx \\ &= \int_{0}^{1} \left[ (1-\frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2-\cdots\right]dx\end{align*}

2. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, I got

\begin{align*} I &= \int_{0}^{1}\left[ 1-\frac{1}{2}(-x)+\frac{1}{3}(-x)^2-\frac{1}{4}(-x)^3+\cdots\right] dx \\ &= \int_{0}^{1}\left[ 1+\frac{1}{2}(x)+\frac{1}{3}(x)^2+\frac{1}{4}(x)^3 +\cdots\right] dx \\ &= (1-0)+\frac{1}{2}\Big( \frac{1}{2} -0\Big) + \frac{1}{3}\Big( \frac{1}{3}-0\Big) + \frac{1}{4}\Big( \frac{1}{4}-0\Big)+\cdots \\ &= \sum_{k=1}^{\infty}\frac{1}{k^2} \\ &= \frac{\pi^2}{6}\end{align*}

7. ## Re: integral

Originally Posted by chisigma
Some time ago I 'discovered' the general formula...

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

... and if we take into account that...

$\frac{1}{1-x}= \sum_{m=0}^{\infty} x^{m}$ (2)

... we obtain...

$\int_{0}^{1} \frac{\ln x}{x-1}\ dx = \sum_{m=0}^{\infty} \frac{1}{(m+1)^{2}}= \frac{\pi^{2}}{6}$
... of course it is not a difficult task to extend the concept arriving, starting from (1), to the more general formula...

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

... where $\zeta(*)$ is the 'Riemann Zeta Function'...

Kind regards

$\chi$ $\sigma$