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

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

• The answer is $\dfrac{2\pi}{4+y^2}$ . Indeed, you have already proven that $F(y)$ is even. – Lucian Feb 11 '14 at 6:26
• Hello,@Lucian,How find this integral? – china math Feb 11 '14 at 6:42
• What about partial fraction decomposition ? – Claude Leibovici Feb 11 '14 at 6:49
• Partial fractions, dear friend. $\dfrac{2x+3y}{(1+[x+y]^2)}-\dfrac{2x-y}{1+x^2}$ – Lucian Feb 11 '14 at 6:53
• Does $y \in {\mathbb R}$ ?. – Felix Marin Sep 13 '14 at 18:30

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.

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.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With the identities: \begin{align} {1 \over 1 + \pars{x + y}^{2}} &=\int_{-\infty}^{\infty}{\delta\pars{x + y - z} \over 1 + z^{2}}\,\dd z \\[3mm] \delta\pars{x + y - z} &=\int_{-\infty}^{\infty} \expo{\ic k\pars{x\ +\ y\ -\ z}}\,{\dd k \over 2\pi} \end{align} we'll have

\begin{align} {\rm F}\pars{y}& =\color{#66f}{\large% \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} \end{align}

\begin{align} {\rm F}\pars{y}& =\color{#66f}{\large% \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}} \end{align}