Please help me to sole the following..

$\displaystyle I=\int{\int{\ln\left(\frac{1}{\sqrt{x^2+y^2}} \right) dxdy$

Thank you very much..

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- Aug 18th 2013, 07:14 AMkjchauhanDouble Integral
Please help me to sole the following..

$\displaystyle I=\int{\int{\ln\left(\frac{1}{\sqrt{x^2+y^2}} \right) dxdy$

Thank you very much.. - Aug 18th 2013, 11:06 AMHallsofIvyRe: Double Integral
The first thing you should do is use properties of logarithms to write the integrand as $\displaystyle ln\left(\frac{1}{\sqrt{x^2+ y^2}}\right)= -\frac{1}{2}ln(x^2+ y^2)$. then change to polar coordinates so that $\displaystyle x^2+ y^2= r^2$ and $\displaystyle dxdy= r drd\theta$. The integral become $\displaystyle -\frac{1}{2}\int\int ln(r^2)r dr\theta= -\frac{1}{2}\int d\theta \int ln(r^2) r dr$. Can you do that?

- Aug 18th 2013, 06:59 PMkjchauhanRe: Double Integral
Thank you very much.

- Aug 18th 2013, 09:14 PMkjchauhanRe: Double Integral
I got the answer

$\displaystyle I = -\frac{1}{4}\left[ tan^{-1}\left( \frac{y}{x} \right) \left[ \left(x^2+y^2 \right) ln \left(x^2+y^2 \right) - \left(x^2+y^2 \right) + c_{1} \right] + c_{2} \right] $

Is it correct? - Aug 18th 2013, 11:38 PMProve ItRe: Double Integral
What are your bounds? Or what is your region you're integrating over?

- Aug 19th 2013, 06:20 AMHallsofIvyRe: Double Integral
ProveIt: The final integral has both x and y in it, as well as two "constants of integration" so there are no bounds- this is an "indefinite integral".

kjchauhan, I have not calculated it myself but that looks good.