
improper integral
Hi all,
Maybe a stupid question, but how to calculate the following integral:
$\displaystyle \int_{\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}$
I would like to see what is the method of calculating such integrals, because I don't quite get what is written on Wikipedia:
http://upload.wikimedia.org/math/9/7...7152304b68.png
According to the formula above, I assume it's:
$\displaystyle \int_{\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\lim_{t\rig htarrow\infty}\int_{t}^{a}{\frac{dx}{x^2+2x+2}}+\lim_{t\r ightarrow+\infty}\int_{a}^{t}{\frac{dx}{x^2+2x+2}}$
Then I want to use substitution $\displaystyle x+1=u$, but I am not sure how the integration limits will change; is it $\displaystyle a+1$ while $\displaystyle t$ remains intact? If so, then it would be:
$\displaystyle \lim_{t\rightarrow\infty}\int_{t}^{a+1}{\frac{du}{u^2+1}}+\lim_{t\ri ghtarrow+\infty}\int_{a+1}^{t}{\frac{du}{u^2+1}}$
Does it further give something like $\displaystyle \lim_{t\rightarrow\infty}\left(\arctan{(a+1)}\arctan{t}\right)+\lim_{t\rightarrow+\infty}\left( \arctan{t}\arctan{(a+1)}\right)$?
And, finally, $\displaystyle \int_{\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\pi$ :confused:
I have no idea, that was just my guess.... Thanks for any input. (Smile)

It's ok (Wink), you didn't need to change the t because it's going to infinity.
But: $\displaystyle \int_a^b\frac{dx}{x^2+2x+2}=\int_{a+1}^{b+1}\frac{ du}{u^2+1}$