1. ## Improper Riemann Integral

Does the integral $\int\limits_{0}^{1} \frac{\log{x}}{1-x^{2}}$ converge where $x \in [0,1]$.

2. Originally Posted by Chandru1
Does the integral $\int\limits_{0}^{1} \frac{\log{x}}{1-x^{2}}$ converge where $x \in [0,1]$.
Yes. It's value is $\frac{-\pi^2}{8}$. Do you want to see why or do you just want why it's convergent? Well..it doesn't matter until I see that you show some work!

3. Spoiler:

In general is...

$\int x^{n}\cdot \ln x \cdot dx = \frac{x^{n+1}}{n+1}\cdot (\ln x - \frac{1}{n+1}) + c$ (1)

... so that is...

$\int_{0}^{1} x^{n}\cdot \ln x \cdot dx = -\frac {1}{(n+1)^{2}}$ (2)

Using (2) we obtain...

$\int_{0}^{1} \frac{\ln x}{1-x^{2}}\cdot dx= \sum_{n=0}^{\infty} \int_{0}^{1} x^{2n}\cdot \ln x \cdot dx =$

$= - \sum_{n=0}^ {\infty} \frac{1}{(2n+1)^{2}} = - \frac{\pi^{2}}{8}$ (3)

Kind regards

$\chi$ $\sigma$

4. Guys, you're not proving at all that it converges

5. I don't undestand what needs further demostration...

... may be this step?...

$\int_{0}^{1} x^{n}\cdot \ln x \cdot dx = -\frac {1}{(n+1)^{2}}$ (1)

... or may be this?...

$\int_{0}^{1} \frac{\ln x}{1-x^{2}}\cdot dx= \sum_{n=0}^{\infty} \int_{0}^{1} x^{2n}\cdot \ln x \cdot dx =$

$= - \sum_{n=0}^ {\infty} \frac{1}{(2n+1)^{2}} = - \frac{\pi^{2}}{8}$ (2)

Kind regards

$\chi$ $\sigma$

6. Originally Posted by Moo
Guys, you're not proving at all that it converges
I was hoping the OP would do the proof.

7. Originally Posted by Drexel28
I was hoping the OP would do the proof.
If he asks for it, he must have some reason.

chisigma : you're not proving that it converges, you're supposing it does and calculates it as if it was the case.

8. $\int_{1}^{\infty }{\frac{\ln x}{x^{2}-1}\,dx}=\int_{1}^{2}{\frac{\ln x}{x^{2}-1}\,dx}+\int_{2}^{\infty }{\frac{\ln x}{x^{2}-1}\,dx}.$

the first integral converges by limit comparison test with $\int_1^2\frac{\ln x}{x-1}\,dx,$ now put $x\mapsto x+1$ in this integral to get $\int_0^1\frac{\ln(x+1)}{x}\,dx$ which exists since the integrand has limit as $x\to0.$ (1)

for $x\ge2$ we have $\frac{\ln x}{x^{2}-1}\le \frac{2\ln x}{x^{2}},$ and $\int_{2}^{\infty }{\frac{\ln x}{x^{2}}\,dx}<\infty .$ (2)

in your integral put $x\mapsto\frac1x$ and get $-\int_1^\infty\frac{\ln x}{x^2-1}\,dx,$ then by (1) and (2) the conclusion follows.