# improper integral

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• Apr 22nd 2008, 02:22 AM
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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)
• 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: $\displaystyle \int_a^b\frac{dx}{x^2+2x+2}=\int_{a+1}^{b+1}\frac{ du}{u^2+1}$