Question Number 28881 by abdo imad last updated on 31/Jan/18
$${find}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}} \\ $$
Answered by sma3l2996 last updated on 01/Feb/18
![I=∫_(−∞) ^(+∞) (dt/(t^2 +2t+2))=∫_(−∞) ^(+∞) (dt/((t+1)^2 +1)) u=t+1⇒du=dt I=∫_(−∞) ^(+∞) (du/(u^2 +1))=[tan^(−1) (u)]_(−∞) ^(+∞) =(π/2)−(−(π/2))=π](https://www.tinkutara.com/question/Q28895.png)
$${I}=\int_{−\infty} ^{+\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}}=\int_{−\infty} ^{+\infty} \frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${u}={t}+\mathrm{1}\Rightarrow{du}={dt} \\ $$$${I}=\int_{−\infty} ^{+\infty} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\left[{tan}^{−\mathrm{1}} \left({u}\right)\right]_{−\infty} ^{+\infty} =\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)=\pi \\ $$