Calculate the integral$$ \int_0^\infty \frac{x \sin rx }{a^2+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x \sin rx }{a^2+x^2} dx,\quad a,r \in \mathbb{R}. $$ Edit: I was able to solve the integral using complex analysis, and now I want to try and solve it using only real analysis techniques.

  • $\begingroup$ There is also the pole at $-ai$ $\endgroup$ – Ellya Apr 25 '14 at 22:28

It looks like I'm too late but still I wanna join the party. :D

Consider $$ \int_0^\infty \frac{\cos rx}{x^2+a^2}\ dx=\frac{\pi e^{-ar}}{a}. $$ Differentiating the both sides of equation above with respect to $r$ yields $$ \begin{align} \int_0^\infty \frac{d}{dr}\left(\frac{\cos rx}{x^2+a^2}\right)\ dx&=\frac{d}{dr}\left(\frac{\pi e^{-ar}}{a}\right)\\ -\int_0^\infty \frac{x\sin rx}{x^2+a^2}\ dx&=(-a)\frac{\pi e^{-ar}}{a}\\ \Large\int_0^\infty \frac{x\sin rx}{x^2+a^2}\ dx&=\Large\pi e^{-ar}. \end{align} $$ Done! :)

  • $\begingroup$ What is your justification for differentiation under the integral sign? That is, while the left side of the first equation is differentiable with respect to $r$ if the equation is valid, since the right side is differentiable with respect to $r$, how do you know it is valid to compute the $r$-derivative of the left side by the process of differentiation under the integral sign? $\endgroup$ – KCd May 21 '14 at 0:16
  • 2
    $\begingroup$ The first integral was evaluated with a complex contour !!!. The OP is asking for "only real analysis". $\endgroup$ – Felix Marin Jul 17 '14 at 23:28
  • $\begingroup$ I think the second method is a real analysis method @FelixMarin. $\endgroup$ – Tunk-Fey Jul 18 '14 at 9:02

Here is an approach using only real analysis. We will first take $a$ and $r$ both positive.

Under the change of variables $x = au$ we have $$ \int_{-\infty}^\infty \frac{x\sin(rx)}{a^2+x^2}\,dx = \int_{-\infty}^{\infty} \frac{u\sin(ar u)}{1+u^2}\,du $$ and under the change of variables $x = u/r$ we have $$ \int_{-\infty}^\infty \frac{x\sin(rx)}{a^2+x^2}\,dx = \int_{-\infty}^\infty \frac{u\sin u}{(ar)^2+u^2}\,du. $$ Either way, we see the integral depends on $a$ and $r$ only through the value of $ar$ and it'd be simpler to ask about the one-parameter integral $$ \int_{-\infty}^\infty \frac{u\sin(cu)}{1+u^2}\,du = \int_{-\infty}^\infty \frac{u\sin u}{c^2+u^2}\,du $$ for $c > 0$, and then set $c = ar$.

See section 11 of http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf, where it is shown by differentiation under the integral sign (this does not depend on complex analysis) that for $t \geq 0$, $$ \int_{-\infty}^{\infty} \frac{\cos(tx)}{1+x^2}\,dx = \pi e^{-t}. $$ If for $t > 0$ you differentiate both sides with respect to $t$ and can justify differentiation under the integral sign, you'd get $$ \int_{-\infty}^{\infty} \frac{-x\sin(tx)}{1+x^2}\,dx = -\pi e^{-t}, $$ so $$ \int_{-\infty}^{\infty} \frac{x\sin(tx)}{1+x^2}\,dx = \pi e^{-t}. $$ Then for positive $a$ and $r$ we'd get $$ \int_0^\infty \frac{x\sin(rx)}{a^2+x^2}\,dx = \frac{1}{2}\int_{-\infty}^\infty \frac{x\sin(rx)}{a^2+x^2}\,dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{u\sin(aru)}{1+u^2}\,du = \frac{1}{2}\pi e^{-ar}. $$ If $a < 0$ the integral doesn't change, and if $r < 0$ the integral changes its value by $-1$, so for nonzero $a$ and $r$ we'd obtain $$ \int_0^\infty \frac{x\sin(rx)}{a^2+x^2}\,dx = \frac{{\rm sign}(r)}{2}\pi e^{-|ar|}. $$ This remains valid for $r = 0$ if we take ${\rm sign}(0) = 0$. For $a = 0$ the integral is $\int_0^\infty (\sin(rx)/x)\,dx$. For $r > 0$ that is $\int_0^\infty (\sin(x)/x)\,dx$, which is well-known to be $\pi/2$ (see the appendix of my link above for a tedious proof avoiding complex analysis). For $a = 0$ and $r < 0$ the integral is $-\pi/2$. Therefore the displayed formula above is valid for all $a$ and $r$ in $\mathbf R$ provided you can justify differentiation under the integral sign on the left side of $$ \int_{-\infty}^{\infty} \frac{\cos(tx)}{1+x^2}\,dx = \pi e^{-t}. $$ I point out in the link above that the standard hypotheses justifying differentiation under the integral sign do not apply to the integral on the left.



Observe that the integrand is an even function i.e. $f(-x)=f(x)$. Hence notice that your integral is an half of the value of the integral on the whole $\mathbb R$.

Use this and compute the integral of the complex function you wrote on the following path: $\gamma_R=[-R,R],\left[R,R+i\frac a{2}\right], \left[R+i\frac a{2}, -R+i\frac a{2}\right], \left[-R+i\frac a{2},-R\right]$ (say, wlog $a>0$).

Now $g(z)=\frac{ze^{irz}}{a^2+z^2}$ has no pole inside $\gamma_R$, $\forall R>0$ hence we have $\int_{\gamma_R}g(z)\,dz=0$. Then take the limit for $R\rightarrow+\infty$.


Disclaimer: This answer was posted before the OP asked for a proof using only real analysis. See the revision history.

Let $S=\{-ia, ia\}$ and let $ \varphi \colon \Bbb C\setminus S\to \Bbb C, z\mapsto\dfrac{ze^{i(rz)}}{(z^2+a^2)}$.

Given $n\in \Bbb N$ such that $n> a$, define $\gamma (n):=\gamma _1(n)\lor \gamma _2(n)$ with $\gamma _1(n)\colon [-n,n]\to \Bbb C, t\mapsto t$ and $\gamma _2(n)\colon [0,\pi]\to \Bbb C, \theta \mapsto ne^{i\theta}$, ($\gamma (n)$ is an upper semicircle).

Note that $S$ is the set of singularities of $\varphi$ and all of them are simple poles.

I'll be assuming that $r,a\ge0$, the other cases is similar. It follows that $\text{Res}(\varphi, ia)=\left.\dfrac{ze^{irz}}{z+ia}\right|_{z=ia}=\dfrac {e^{-ar}}2$, which is what you got. It doesn't matter what $\text{Res}(\varphi, -ia)$ is.

Quick considerations about winding numbers, inside and outside region of $\gamma(n)$, the fact that $n>a$, the fact that $\varphi$ is holomorphic and the residue theorem yield $$\displaystyle \int \limits_{\gamma (n)}\varphi=2\pi i\cdot \dfrac {e^{-ar}}2=\pi e^{-ar}i.$$

On the other hand $$\int \limits_{\gamma (n)}\varphi=\int \limits _{\gamma _1(n)}\varphi +\int \limits_{\gamma _2(n)}\varphi \tag I$$

Note that $$\displaystyle\int \limits _{\gamma _1(n)}\varphi=\int \limits _{-n}^n\varphi (t)dt=\int \limits_{-n}^n\dfrac{te^{i(rt)}}{t^2+a^2}\mathrm dt=\int \limits_{-n}^n\dfrac{t\cos(rt)+it\sin (rt)}{t^2+a^2}\mathrm dt=i\int \limits _{-n}^n\dfrac{t\sin (rt)}{t^2+a^2}\mathrm dt.$$ The last equality is due to $t\mapsto \dfrac{t\cos (rt)}{t^2+a^2}$ being an odd function and due to the integral being computed on a symmetric interval.

It can be proven that $\displaystyle \lim \limits_{n\to +\infty}\left(\int _{\gamma _2(n)}\varphi\right)=0$, I won't prove this unless you're having trouble with it. In any case, Zaid just brought to my attention this a consequence of a known result, namely Jordan's Lemma.

Thus, taking the limit in $(\text{I})$ yields $\displaystyle \pi e^{-ar}i=i\int \limits _{-\infty}^\infty\dfrac{t\sin (rt)}{t^2+a^2}\mathrm dt.$

  • $\begingroup$ It follows from Jordan lemma that the integral along the circle goes to 0. $\endgroup$ – Zaid Alyafeai Apr 25 '14 at 23:00
  • $\begingroup$ I just googled it. I always prove this bit by hand -_- Thanks. $\endgroup$ – Git Gud Apr 25 '14 at 23:01
  • $\begingroup$ The critical part is to know that $\sin(x)\geq \frac{2}{\pi}x$ on $[0,\pi/2]$. $\endgroup$ – Zaid Alyafeai Apr 25 '14 at 23:04
  • $\begingroup$ Thanks! I independently came to the same result using this technique. However, I am more interested in whether one can arrive at this answer without complex integration. Can you think of a way to solve it using only real analysis? $\endgroup$ – user99026 Apr 25 '14 at 23:12
  • $\begingroup$ @Kalashnik Hold on. I'll try something. No promises though. I just noticed a small mistake in my answer. I implicitly assumed that $a\ge 0$ when considering the winding numbers of the poles. If $a<0$ it is similar. $\endgroup$ – Git Gud Apr 25 '14 at 23:13

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x\sin\pars{rx} \over a^{2} + x^{2}}\,\dd x:\ {\large ?}}$.

It's $\large\tt\underline{possible}$ to evaluate the integral $\ds{\large\tt\mbox{without a Complex Contour !!!}}$

Note that \begin{align} &\color{#c00000}{\int_{0}^{\infty}{x\sin\pars{rx} \over a^{2} + x^{2}}\,\dd x} =-\,\partiald{}{r}\int_{0}^{\infty}{\cos\pars{rx} \over a^{2} + x^{2}}\,\dd x =-\,\half\,\Re\partiald{\fermi\pars{r}}{r} \\[3mm]&\mbox{where}\quad\fermi\pars{r}\equiv \int_{-\infty}^{\infty}{\expo{\ic rx} \over a^{2} + x^{2}}\,\dd x\,,\qquad \fermi\pars{0} = {\pi \over \verts{a}}\tag{1} \end{align}

\begin{align} &\partiald[2]{\fermi\pars{r}}{r}\equiv \int_{-\infty}^{\infty}{-x^{2}\expo{\ic rx} \over a^{2} + x^{2}}\,\dd x =-2\pi\,\delta\pars{r} + a^{2}\fermi\pars{r}\,,\ \imp\ \pars{\partiald[2]{}{r} - a^{2}}\fermi\pars{r} = -2\pi\,\delta\pars{r} \end{align}

Then, $$ \left.\fermi\pars{r}\right\vert_{\,r\ <\ 0}={\pi \over \verts{a}}\expo{\verts{a}r}\,,\qquad \left.\fermi\pars{r}\right\vert_{\,r\ >\ 0}={\pi \over \verts{a}}\,\expo{-\verts{a}r} $$ which is continuous at $\ds{r = 0}$ and satisfies $\ds{\fermi'\pars{0^{+}} - \fermi'\pars{0^{-\vphantom{+}}} = -2\pi}$

$$ \fermi\pars{r} = {\pi \over \verts{a}}\,\expo{-\verts{a}\verts{r}} $$

$$\color{#66f}{\large% \int_{0}^{\infty}{x\sin\pars{rx} \over a^{2} + x^{2}}\,\dd x} =-\,\half\,\partiald{}{r}\pars{{\pi \over \verts{a}}\,\expo{-\verts{a}\verts{r}}} =\color{#66f}{\large\half\pi\,\sgn\pars{r}\expo{-\verts{ar}}} $$


We can use the following known results to evaluate $$ \int_0^\infty e^{-xt}\cos (at)dt=\frac{x}{a^2+x^2}, \int_0^\infty\frac{\cos(\pi x)}{r^2+x^2}dx=\frac{\pi}{2re^{ar}}. $$ So \begin{eqnarray} I&=&2\int_0^\infty \frac{x\sin(rx)}{a^2+x^2}dx=\int_0^\infty \sin(r x) \left(\int_0^\infty e^{-xt}\cos(at)dt\right)dx\\ &=&\int_0^\infty\cos(at)\left(\int_0^\infty e^{-xt}\sin(rx)dtx\right)dt\\ &=&r\int_0^\infty\frac{\cos(a t)}{r^2+t^2}dt\\ &=&\frac{\pi}{2e^{ar}}. \end{eqnarray}


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