# improper integral

• Apr 22nd 2008, 02:22 AM
disclaimer
improper integral
Hi all,

Maybe a stupid question, but how to calculate the following integral:
$\int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}$
I would like to see what is the method of calculating such integrals, because I don't quite get what is written on Wikipedia:
According to the formula above, I assume it's:
$\int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\lim_{t\rig htarrow-\infty}\int_{t}^{a}{\frac{dx}{x^2+2x+2}}+\lim_{t\r ightarrow+\infty}\int_{a}^{t}{\frac{dx}{x^2+2x+2}}$
Then I want to use substitution $x+1=u$, but I am not sure how the integration limits will change; is it $a+1$ while $t$ remains intact? If so, then it would be:
$\lim_{t\rightarrow-\infty}\int_{t}^{a+1}{\frac{du}{u^2+1}}+\lim_{t\ri ghtarrow+\infty}\int_{a+1}^{t}{\frac{du}{u^2+1}}$
Does it further give something like $\lim_{t\rightarrow-\infty}\left(\arctan{(a+1)}-\arctan{t}\right)+\lim_{t\rightarrow+\infty}\left( \arctan{t}-\arctan{(a+1)}\right)$?
And, finally, $\int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\pi$ :confused:
I have no idea, that was just my guess.... Thanks for any input. (Smile)
• Apr 22nd 2008, 02:53 AM
PaulRS
It's ok (Wink), you didn't need to change the t because it's going to infinity.

But: $\int_a^b\frac{dx}{x^2+2x+2}=\int_{a+1}^{b+1}\frac{ du}{u^2+1}$