Integral help needed

**kevinlightman** · August 30th 2009, 03:23 AM

$I_{n} = \int_{0}^{\infty}\frac{ln(x)}{n^2 + x^2}dx$

I tried going by parts and using arctan but I get stuck after there, can anyone point me in the right direction please?

**luobo** · August 30th 2009, 05:08 AM

Originally Posted by kevinlightman

$I_{n} = \int_{0}^{\infty}\frac{\ln x}{n^2 + x^2}dx$

I tried going by parts and using arctan but I get stuck after there, can anyone point me in the right direction please?

(Step 1) $\text{ Let } t=\frac{n}{x} \Rightarrow I_n = \frac{1}{n} \int_0^\infty \frac{\ln n - \ln t} {1 + t^2} \; dt=\frac{\pi \ln n}{2n} - \frac{1}{n}\int_0^\infty \frac{ \ln t} {1 + t^2} \; dt$

(Step 2) $\int_0^\infty \frac{ \ln t} {1 + t^2} \; dt = \int_0^1 \frac{ \ln t} {1 + t^2} \; dt + \int_1^\infty \frac{ \ln t} {1 + t^2} \; dt$ $= \int_0^1 \frac{ \ln t} {1 + t^2} \; dt - \int_0^1 \frac{ \ln t} {1 + t^2} \; dt = 0$

(Step 3) $I_n = \frac{\pi \ln n}{2n}$

**Krizalid** · August 30th 2009, 05:11 AM

Substitution $x=nt$ will solve the problem.

(Ahhh, luobo beat me to it.)

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