Find this integral $$F(y)=\int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)(1+(x+y)^2)}$$
my try: since $$F(-y)=\int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)(1+(x-y)^2)}$$ let $x=-u$,then $$F(-y)=\int_{-\infty}^{+\infty}\dfrac{dx}{(1+x^2)(1+(-x-y)^2)}=F(y)$$
But I can't find $F(y)$,Thank you
On
An easy way to evaluate the above integral is using the Parseval's theorem:
Let $f(x)$ and $g(x)$ be integrable, and let $\hat{f}(\xi)$ and $\hat{g}(\xi)$ be their Fourier transforms. If $f(x)$ and $g(x)$ are also square-integrable, then we have Parseval's theorem (Rudin 1987, p. 187): $$\int_{-\infty}^{\infty} f(x) \overline{g(x)} \,{\rm d}x = \int_{-\infty}^\infty \hat{f}(\xi) \overline{\hat{g}(\xi)} \,d\xi,$$ where the bar denotes complex conjugation.
You need the following fact:
$$ \mathcal{F} \left\{\frac{1}{1+(x+y)^2} \right\}(w) =\pi e^{iyw}e^{-|w|}. $$
Check the different conventions of the Fourier transform and its tables.
On
How about using the residue theorem? We have $$F(y)= 2\pi i \sum_{x^*} \mathop{\rm Res}_{x=x^*} f(x),$$ where the sum ranges over all the poles of the integrand $$f(x)= \frac1{(1+x^2)[1+(x+y)^2]}$$ in the upper half-plane. These poles are situated at $x^*= i, i-y$ (assuming $y$ to be real).
As the poles are simple poles, we obtain the residues by $$ \mathop{\rm Res}_{x=x^*} f(x) =\lim_{x\to x^*} (x-x^*)f(x) .$$ Thus, we have $$ \mathop{\rm Res}_{x=i} f(x) = \frac{1}{2 i [1+(i + y)^2]}$$ and $$ \mathop{\rm Res}_{x=i-y} f(x) = \frac{1}{2 i [1+(i-y)^2]}.$$
So, we obtain $$ F(y) = \pi \left[\frac{1}{1+(i + y)^2} + \frac{1}{1+(i-y)^2} \right] = \frac{2 \pi}{4+y^2}$$ as the final result.
Hint
Use first partial fraction decompositions. Then, integrate between $-a$ and $+a$. You should arrive to something like $$\frac{-\log \left((a-y)^2+1\right)+\log \left((a+y)^2+1\right)+y \left(\tan ^{-1}(a-y)+\tan ^{-1}(a+y)+2 \tan ^{-1}(a)\right)}{y \left(y^2+4\right)}$$ Simplify as much as you can and go to limits. You will arrive to Lucian's solution.