Without using Integral by Part

**LordoftheFlies** · January 1st 2013, 10:19 AM

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

Now switch the order of integration:

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

**uniquesailor** · January 1st 2013, 09:45 PM

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?

**LordoftheFlies** · January 2nd 2013, 02:38 AM

It's a standard trick. Watch this video, he'll explain the basic idea better than I can.

Spoiler:

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