How find this integral $F(y)=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)(1+(x+y)^2)}$

Find this integral

my try: since
let $x=-u$,then

But I can’t find $F(y)$,Thank you

Solutions Collecting From Web of "How find this integral $F(y)=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)(1+(x+y)^2)}$"

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
$$ \mathop{\rm Res}_{x=i} f(x) = \frac{1}{2 i [1+(i + y)^2]}$$
$$ \mathop{\rm Res}_{x=i-y} f(x) = \frac{1}{2 i [1+(i-y)^2]}.$$

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.


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.

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.

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With the identities:
{1 \over 1 + \pars{x + y}^{2}}
&=\int_{-\infty}^{\infty}{\delta\pars{x + y – z} \over 1 + z^{2}}\,\dd z
\delta\pars{x + y – z} &=\int_{-\infty}^{\infty}
\expo{\ic k\pars{x\ +\ y\ -\ z}}\,{\dd k \over 2\pi}
we’ll have

{\rm F}\pars{y}&
\int_{-\infty}^{\infty}{\dd x \over \pars{1 + x^{2}}\bracks{1 + \pars{x + y}^{2}}}}
\\[8mm]&=\int_{-\infty}^{\infty}\expo{\ic ky}
\pars{\int_{-\infty}^{\infty}{\expo{\ic kx} \over 1 + x^{2}}\,\dd x}
\pars{\int_{-\infty}^{\infty}{\expo{-\ic kz} \over 1 + z^{2}}\,\dd z}
\,{\dd k \over 2\pi}
\\[8mm]&=\int_{-\infty}^{\infty}\expo{\ic ky}
\verts{\int_{-\infty}^{\infty}{\expo{\ic kx} \over 1 + x^{2}}\,\dd x}^{2}
\,{\dd k \over 2\pi}
=\int_{-\infty}^{\infty}\expo{\ic ky}
\verts{2\pi\ic\,{\expo{\ic\verts{k}\ic} \over 2\ic}}^{2}\,{\dd k \over 2\pi}
\\[8mm]&=\half\,\pi\int_{-\infty}^{\infty}\expo{\ic ky}\expo{-2\verts{k}}\,\dd k
=\half\,\pi\int_{-\infty}^{\infty}\cos\pars{ky}\expo{-2\verts{k}}\,\dd k
=\pi\,\Re\int_{0}^{\infty}\expo{\pars{-2 + y\ic}k}\,\dd k
\\[8mm]&=\pi\,\Re\pars{1 \over 2 – y\ic}

{\rm F}\pars{y}&
\int_{-\infty}^{\infty}{\dd x \over \pars{1 + x^{2}}\bracks{1 + \pars{x + y}^{2}}}}
=\color{#66f}{\large{2\pi \over y^{2} + 4}}