# Without using Integral by Part

Printable View

• Dec 10th 2012, 10:30 AM
uniquesailor
Without using Integral by Part
• Jan 1st 2013, 11:19 AM
LordoftheFlies
Re: Without using integral by parts
$\displaystyle\int_0^{\infty}e^{-t}\arctan t\;dt=\int_0^{\infty}\int_0^t \frac{e^{-t}}{1+x^2} \;dx\;dt$

Now switch the order of integration:

$\qquad\qquad=\int_0^{\infty}\int_x^{\infty} \frac{e^{-t}}{1+x^2} \;dt\;dx=\int_0^{\infty}\frac{e^{-x}}{1+x^2}\;dx$
• Jan 1st 2013, 10:45 PM
uniquesailor
Re: Without using integral by parts
Quote:

Originally Posted by LordoftheFlies
$\displaystyle\int_0^{\infty}e^{-t}\arctan t\;dt=\int_0^{\infty}\int_0^t \frac{e^{-t}}{1+x^2} \;dx\;dt$

Now switch the order of integration:

$\qquad\qquad=\int_0^{\infty}\int_x^{\infty} \frac{e^{-t}}{1+x^2} \;dt\;dx=\int_0^{\infty}\frac{e^{-x}}{1+x^2}\;dx$

Can u explain that step?

https://sites.google.com/site/atzahr...helpforum1.png
• Jan 2nd 2013, 03:38 AM
LordoftheFlies
Re: Without using Integral by Part
It's a standard trick. Watch this video, he'll explain the basic idea better than I can.

Spoiler: