improper integrals

**Rine198** · June 4th 2011, 09:23 PM

Find

$\lim_{R\to \infty }\int ^a_b \frac{x}{1+x^2}$

where a=2R and b=-R

please help me with this, i manage to integrate it, but i don't how to find the limit , the answer in ln(2)

thanks

**Chris L T521** · June 4th 2011, 09:32 PM

Originally Posted by Rine198

Find

$\lim_{R\to \infty }\int ^a_b \frac{x}{1+x^2}$

where a=2R and b=-R

please help me with this, i manage to integrate it, but i don't how to find the limit , the answer in ln(2)

thanks

Spoiler:

Can you proceed?

**Rine198** · June 4th 2011, 11:03 PM

how do i explain that :

$\int ^a_b \frac{x}{1+x^2 }$

where $a=\infty$ and $b=-\infty$ is divergent?

thanks

**chisigma** · June 4th 2011, 11:34 PM

Originally Posted by Rine198

how do i explain that :

$\int ^a_b \frac{x}{1+x^2 }$

where $a=\infty$ and $b=-\infty$ is divergent?

thanks

The question is controversial... pratically the integral ...

$\int_{-\infty}^{+ \infty} \frac{x}{1+x^{2}}\ dx = \lim_{a \rightarrow \infty, b \rightarrow -\infty} \ln \sqrt{\frac{1+a^{2}}{1+b^{2}}$ (1)

... doesn't exist if a and b are mutually independent. The integral...

$\text{PV} \int_{-\infty}^{+ \infty} \frac{x}{1+x^{2}}\ dx = \lim_{a \rightarrow \infty} \ln \sqrt{\frac{1+a^{2}}{1+a^{2}}$ (2)

... where PV means 'principal value' is the (1) with a=b and of course it exists and is zero...

Kind regards

$\chi$ $\sigma$

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