improper integral

**disclaimer** · April 22nd 2008, 02:22 AM

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, $\int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\pi$

I have no idea, that was just my guess.... Thanks for any input.

**PaulRS** · April 22nd 2008, 02:53 AM

It's ok

, you didn't need to change the t because it's going to infinity.

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

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