# Without using Integral by Part

• December 10th 2012, 09:30 AM
uniquesailor
Without using Integral by Part
• January 1st 2013, 10:19 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:

$\qquad\qquad=\int_0^{\infty}\int_x^{\infty} \frac{e^{-t}}{1+x^2} \;dt\;dx=\int_0^{\infty}\frac{e^{-x}}{1+x^2}\;dx$
• January 1st 2013, 09:45 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?