cauchy-integral-formulacomplex-analysisintegrationresidue-calculus
I have to evaluate the following integral for $a>0$:
$$\int_0^\infty \left( \frac{\sin az}{z^2+1}\right)^2 dz$$
I don't exactly know how to do this kind of integral. But I think I need to use the residue theorem. Maybe I could use the trick $\cos(2a) = 1 – \sin^2(a)$, but that makes the integral much more difficult I think.
You have found, correctly, that $$\begin{eqnarray*} I &\equiv& \int_0^\infty\frac{x^{1/3}dx}{1+x^2} \\ &=& \frac{2}{1-e^{2\pi i/3}} \int_{-\infty}^\infty d z\, \frac{z^{5/3}}{1+z^4}. \end{eqnarray*}$$ The second integral has a branch cut at $z=0$ and singularities at the roots of $1+z^4$. Close the contour in the upper half-plane. Let $\gamma$ denote the contour. We will pick up residues at $e^{i\pi/4}$ and $e^{i 3\pi/4}$.
We deform the contour slightly near $z=0$ to avoid the cut. Letting $z=\epsilon e^{i\theta}$ we see that the contribution to the integral here goes like $\epsilon^{1+5/3} = \epsilon^{8/3}$ and so vanishes in the limit.
Letting $z = R e^{i\theta}$, we find the integral over the semicircle at infinity goes like $R^{1+5/3-4} = 1/R^{4/3}$. Thus, the contribution from this part of the contour is also zero. Therefore, $\int_{-\infty}^\infty d z\, z^{5/3}/(1+z^4) = \int_\gamma d z\, z^{5/3}/(1+z^4)$.
We find $$\begin{eqnarray*} I &=& \frac{2}{1-e^{2\pi i/3}} \int_\gamma d z\, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{2}{1-e^{2\pi i/3}} 2\pi i \sum\mathrm{Res} \, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{\pi}{\sqrt{3}} \end{eqnarray*}$$ where the residues are to be taken from the upper half-plane.
The residue of $f(z)$ at $z=z_0$ is just the coefficient of $(z-z_0)^{-1}$ in the Laurent series and, of course, $z^{5/3}/(1+z^4)$ has a Laurent series about the zeros of $1+z^4$. (It does not, however, have a Laurent series about $z=0$.)
A simpler solution
Recognize that the integral $I$ is already in the form $$\int_0^\infty d t\, t^\beta f(t^2).$$ But $$\begin{eqnarray*} \int_0^\infty d t\, t^\beta f(t^2) &=& \frac{1}{1+e^{\beta \pi i}} \int_{-\infty}^\infty d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} \int_\gamma d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} 2\pi i \sum \mathrm{Res} \, t^\beta f(t^2) \end{eqnarray*}$$ assuming the contributions from the contour around the branch cut at $z=0$ and the contour at infinity vanish. (Again, we have closed the contour in the upper half-plane and pick up only the residues residing there.) It is also important that the singularities of $f$ do not lie on the real line.
The contribution near $z=0$ goes like $\epsilon^{1+1/3} = \epsilon^{4/3}$. On the semicircle at infinity the integral goes like $R^{1+1/3-2} = R^{-2/3}$. Thus, $\int_{-\infty}^\infty d t\, t^{1/3}/(1+t^2) = \int_\gamma d t\, t^{1/3}/(1+t^2)$.
There is one residue, at $t=i=e^{i\pi/2}$, so we have halved our work. We find $$\begin{eqnarray*} I &=& \frac{1}{1+e^{\pi i/3}} 2\pi i \frac{e^{i\pi/6}}{2i} \\ &=& \frac{\pi}{2\cos \pi/6} \\ &=& \frac{\pi}{\sqrt{3}}. \end{eqnarray*}$$
I'm pretty sure I remember doing this one before...
Ah, there it is. AoPS link.
Quoting myself:
Well, you can't just leave it there. This one definitely deserves more than just an answer.
Now, trying to work out an actual solution - $1$ is not a special point for $\sin$. There's nothing special about picking $\frac1x$ there - so, to give ourselves more handles to work with, we stretch the problem. Define $$F(a) = \int_0^{\infty}\sin(x)\sin\left(\frac{a}{x}\right)\,dx$$ Now, we would like to differentiate $F$. Differentiating under the integral sign gets us $F'(a) \stackrel{?}{=} \int_0^{\infty}\frac1x\sin(x)\cos\left(\frac{a}{x}\right)\,dx$ and then $F''(a) \stackrel{?}{=} \int_0^{\infty}\frac{-1}{x^2}\sin(x)\sin\left(\frac{a}{x}\right)\,dx$, while substituting $t=\frac{a}x$ gets us $F(a) = \int_{\infty}^{0}\sin\left(\frac{a}{t}\right)\sin(t)\cdot\frac{-a}{t^2}\,dt \stackrel{?}{=} -aF''(a)$.
Why the question marks? Because I'm doing things with conditionally convergent integrals that I would need absolute convergence to justify.
It's still informative. That's not the Bessel differential equation, but it's related - looking at the power series, we would get a Bessel function of $2\sqrt{a}$. In light of that, let's redefine: $$G(a) = \int_0^{\infty}\sin(x)\sin\left(\frac{a^2}{x}\right)\,dx$$ Now, we need to massage this into a form that will let us justify the differentiation. First, split at $x=a$ and fold with that substitution: \begin{align*}G(a) &= \int_0^a \sin(x)\sin\left(\frac{a^2}{x}\right)\,dx + \int_a^{\infty} \sin(x)\sin\left(\frac{a^2}{x}\right)\,dx\\ &= \int_0^a \sin(x)\sin\left(\frac{a^2}{x}\right)\,dx + \int_a^0 \sin\left(\frac{a^2}{t}\right)\sin(t)\cdot\frac{-a^2}{t^2}\,dt\\ &= \int_0^a \sin(x)\sin\left(\frac{a^2}{x}\right)\,dx + \int_0^a \frac{a^2}{x^2}\sin(x)\sin\left(\frac{a^2}{x}\right)\,dx\\ G(a) &= \int_0^a \left(1+\frac{a^2}{x^2}\right)\sin(x)\sin\left(\frac{a^2}{x}\right)\,dx\end{align*} This form is improper at zero due to the rapid growth (balanced by oscillation), so we integrate by parts: \begin{align*}G(a) &= \int_0^a \left(1+\frac{a^2}{x^2}\right)\sin(x)\sin\left(\frac{a^2}{x}\right)\,dx\\ &\,_{dv = \tfrac{a^2}{x^2}\sin\tfrac{a^2}{x}\,dx,\quad v = \cos\tfrac{a^2}{x}}^{u = (\tfrac{x^2}{a^2}+1)\sin x,\quad du = (\tfrac{x^2}{a^2}+1)\cos x + \tfrac{2x}{a^2}\sin x\,dx}\\ G(a) &= \left[\left(\frac{x^2}{a^2}+1\right)\sin x\cos\frac{a^2}{x}\right]_{x=0}^{x=a} - \int_0^a \cos\frac{a^2}{x}\left(\left(\frac{x^2}{a^2}+1\right)\cos x + \frac{2x}{a^2}\sin x\right)\,dx\\ &= \sin(2a) - \int_0^a \cos\frac{a^2}{x}\left(\left(\frac{x^2}{a^2}+1\right)\cos x + \frac{2x}{a^2}\sin x\right)\,dx\end{align*} This form is now absolutely convergent; it's the integral of a bounded function on a bounded interval, continuous except at one endpoint. If we differentiate it - well, then we get terms in which we multiply by $\frac{2a}{x}$, and it's improper again. Only the $\cos\frac{a^2}{x}\cos x$ term causes trouble, so let's separate that out: $$G(a) = \sin(2a) - \int_0^a \cos\frac{a^2}{x}\left(\frac{x^2}{a^2}\cos x + \frac{2x}{a^2}\sin x\right)\,dx - \int_0^a \cos\frac{a^2}{x}\cos x\,dx$$ \begin{align*}\int_0^a \cos\frac{a^2}{x}\cos x\,dx &= \left[-\frac{x^2}{a^2}\cos x\sin\frac{a^2}{x}\right]_{x=0}^{x=a} + \int_0^a \sin\frac{a^2}{x}\left(\frac{2x}{a^2}\cos x - \frac{x^2}{a^2}\sin(x)\right)\,dx\\ G(a) &= \frac32\sin(2a) -\int_0^a \frac{x^2}{a^2}\cos\left(x+\frac{a^2}{x}\right) + \frac{2x}{a^2}\sin\left(x+\frac{a^2}{x}\right)\,dx\end{align*} That's a lot of trig identities that go into that last line - mostly the angle-sum identities, which our manipulation was nice enough to set up for us cleanly.And now, we can finally differentiate cleanly. \begin{align*}G(a) &= \frac32\sin(2a) -\int_0^a \frac{x^2}{a^2}\cos\left(x+\frac{a^2}{x}\right) + \frac{2x}{a^2}\sin\left(x+\frac{a^2}{x}\right)\,dx\\ G'(a) &= 3\cos(2a) - \cos(2a) - \frac{2}{a}\sin(2a)\\ &\quad+\int_0^a \frac{2x^2}{a^3}\cos\left(*\right) + \frac{2x}{a}\sin\left(*\right) + \frac{4x}{a^3}\sin\left(*\right) - \frac{4}{a}\cos\left(*\right)\,dx\\ G'(a) &= 2\cos(2a) -\frac{2}{a}\sin(2a)\\ &\quad + \int_0^a \left(\frac{2x^2}{a^3}-\frac{4}{a}\right)\cos\left(x+\frac{a^2}{x}\right) + \left(\frac{2x}{a} + \frac{4x}{a^3}\right)\sin\left(x+\frac{a^2}{x}\right)\,dx\end{align*} And now... we'd like to differentiate again, but one of the terms doesn't have any powers of $x$ to cushion it, and differentiating it will break absolute convergence. Separate it out and integrate by parts again: \begin{align*}\int_0^a \frac{x^2}{a^3}\left(1-\frac{a^2}{x^2}\right)\cos\left(x+\frac{a^2}{x}\right)\,dx &= \left[\frac{x^2}{a^3}\sin\left(x+\frac{a^2}{x}\right)\right]_{x=0}^{x=a} - \int_0^a \frac{2x}{a^3}\sin\left(x+\frac{a^2}{x}\right)\,dx\\ &= \frac{1}{a}\sin(2a) - \int_0^a \frac{2x}{a^3}\sin\left(x+\frac{a^2}{x}\right)\,dx\end{align*} Multiply by $4$ and add/subtract, to clear the $\frac{4}{a}\cos(*)$ term: \begin{align*} G'(a) &= 2\cos(2a) + \frac{2}{a}\sin(2a) + \int_0^a -\frac{2x^2}{a^3}\cos\left(x+\frac{a^2}{x}\right) + \left(\frac{2x}{a} - \frac{4x}{a^3}\right)\sin\left(x+\frac{a^2}{x}\right)\,dx\\ G'(a) &= 2\cos(2a) - \frac1a\sin(2a) + \frac2a G(a) + \int_0^a \frac{2x}{a}\sin\left(x+\frac{a^2}{x}\right)\,dx\\ G''(a) &= -4\sin(2a) - \frac2a\cos(2a) + \frac1{a^2}\sin(2a) -\frac2{a^2} G(a) + \frac2a G'(a) + 2\sin(2a)\\ &\quad + \int_0^a 4\cos(*) - \frac{2x}{a^2}\sin(*)\,dx\\ G''(a) &= -2\sin(2a) + \frac1{a^2}\sin(2a) - \frac2a\cos(2a) -\frac2{a^2} G(a) + \frac2a G'(a)\\ &\quad + \int_0^a 4\cos(*) - \frac{2x}{a^2}\sin(*)\,dx\end{align*} Now, we deal with those integrals. From an integration by parts earlier, $\int_0^a (x^2-a^2)\cos(*) + 2x\sin(*)\,dx = a^2\sin(2a)$. From our expression for $G$, $\int_0^a x^2\cos(*) + 2x\sin(*) = \frac{3a^2}{2}\sin(2a)-a^2 G(a)$. Subtract the two, divide by $a^2$, and $\int_0^a \cos(*)\,dx = \frac12\sin(2a) - G(a)$. Similarly, from our latest expression for $G'$, $\int_0^a x\sin(*)\,dx = \frac{a}{2}G'(a) - G(a) - a\cos(2a) + \frac12\sin(2a)$. Apply these to the formula for $G''$, and \begin{align*}G''(a) &= -2\sin(2a) + \frac1{a^2}\sin(2a) - \frac2a\cos(2a) - \frac2{a^2}G(a) + \frac2{a}G'(a) +\\ &\quad 2\sin(2a) - 4 G(a) - \frac1a G'(a) + \frac2{a^2}G(a) + \frac2a\cos(2a) - \frac1{a^2}\sin(2a)\\ G''(a) &= \frac1{a}G'(a) - 4 G(a) \end{align*} This almost looks like a stretched Bessel equation of order zero - but the middle term has the wrong sign, and that's not something we can change by just flipping the sign of $x$ or something. No, the way to change that middle term is to multiply the function by powers of $x$. Combining with that stretch, let $H(t)=\frac1t G\left(\frac{t}{2}\right)$. Then $H'(t)=\frac{1}{2t}G'\left(\frac{t}{2}\right)-\frac1{t^2}G\left(\frac{t}{2}\right)$, $H''(t)=\frac{1}{4t}G''\left(\frac{t}{2}\right)-\frac1{t^2}G'\left(\frac{t}{2}\right)+\frac{2}{t^3}G\left(\frac{t}{2}\right)$, and \begin{align*}t^2 H''(t) + t H(t) + (t^2-1)H(t) &= \frac{t}{4}G''\left(\frac{t}{2}\right)-G'\left(\frac{t}{2}\right)+\frac2t G\left(\frac{t}{2}\right) + \frac12G'\left(\frac{t}{2}\right)\\ &\quad - \frac1t G\left(\frac{t}{2}\right) + t G\left(\frac{t}{2}\right) - \frac1t G\left(\frac{t}{2}\right)\\ &= \frac{t}{4}G''\left(\frac{t}{2}\right) - \frac12 G'\left(\frac{t}{2}\right) + tG\left(\frac{t}{2}\right)\\ &\stackrel{s=t/2}{=} \frac{s}{2}\left(G''(s) -\frac1s G'(s) + 4 G(s)\right) = 0\end{align*} That equation for $H(t)$ is the Bessel equation of order $1$.
Now, we still need an initial condition. What is $G(0)$? Well, our equation degenerates there in several ways, but we can take a limit: \begin{align*}\lim_{a\to 0^+}G(a) &= \lim_{a\to 0^+}\frac32\sin(2a) -\lim_{a\to 0^+}\int_0^a \frac{x^2}{a^2}\cos\left(x+\frac{a^2}{x}\right) + \frac{2x}{a^2}\sin\left(x+\frac{a^2}{x}\right)\,dx\\ &= 0 - \lim_{a\to 0^+}\int_0^1 \left(t^2\cos\left(at+\frac{a}{t}\right) + \frac{2t}{a}\sin\left(at+\frac{a}{t}\right)\right)\cdot a\,dt = 0\end{align*} The first part of that integral is bounded by $at^2$, which tends to zero uniformly. For the second part, we use $|\sin(x)|\le |x|$; it's bounded by $2at^2 + 2a$, which also goes to zero uniformly. But that's not enough - we're going to divide by $x$ at least once. We go one step stronger, and look at $\lim_{a\to 0^+}\frac1a G(a)$. The $\sin$ term gets us $3$. In the integral term, the integrand tends to $t^2\cos(0) + 2t^2+2$ pointwise; by dominated convergence (using the same estimates as before), $\lim_{a\to 0^+}\frac1a G(a) = 3 - 3 = 0$. That... well, that tells us $H$ is a multiple of the standard Bessel function $J_1$. We need to go another degree higher to find what multiple. For that, the expression we're working with isn't terribly convenient, so we go back to a much earlier one: \begin{align*}G(a) &= \int_0^a \left(1+\frac{a^2}{x^2}\right)\sin(x)\sin\left(\frac{a^2}{x}\right)\,dx\\ \frac1{a^2}G(a) &= \int_0^1 \left(\frac1{a^2}+\frac1{a^2t^2}\right)\sin(at)\sin\frac{a}{t}\cdot a\,dt\\ \frac1{a^2}G(a) &\approx \int_0^1 \frac{t^2+1}{at^2}\cdot at\cdot \sin\frac{a}{t}\,dt\\ \frac1{a^2}G(a) &\approx \int_0^1 \left(t+\frac1t\right)\sin\frac{a}{t}\,dt\end{align*} As $a\to 0^+$, that last form of the integrand tends to zero. The $t\sin\frac{a}{t}$ term tends to zero uniformly, while the other term is large enough to matter at zero. Now, we substitute: \begin{align*}\frac1{a^2}G(a) &\approx \int_0^1 \frac1t\sin\frac{a}{t}\,dt\\ &\,^{t=\tfrac{a}{x}}_{dt=-\tfrac{a}{x^2}\,dx}\\ \frac1{a^2}G(a) &\approx \int_{\infty}^{a} -\frac{x}{a}\sin(x)\cdot\frac{a}{x^2}\,dx\\ \frac1{a^2}G(a) &\approx \int_a^{\infty} \frac{\sin x}{x}\,dx = \frac{\pi}{2}\end{align*} That's our limit - $\lim_{a\to 0^+} \frac1{a^2}G(a)=\frac{\pi}{2}$. For normalization, note that $J_1(t)\approx \frac{t}{2}$ for $t$ near zero. Then $H(2a)\approx \frac1{2a}G(a) \approx \frac{\pi}{4}a$, and thus $H(2a)=\frac{\pi}{4}J_1(2a)$. Converting back to $G$, $G(a) = 2a H(2a) = \frac{\pi a}{2}J_1(2a)$. The original question asked was $G(1)$, for an integral of $\frac{\pi}{2} J_1(2)$. The answer is confirmed... and it only took me most of a day to get there.
Doing it a second time? It took longer to find the old post than to reedit it for this site's formatting.
Best Answer
The integral can be worked out without the residual theorem. Note
\begin{align} & I(a)=\int_0^\infty \left( \frac{\sin az}{z^2+1}\right)^2 dz\\ & I’(a)=\int_0^\infty \frac{z\sin(2az)}{(z^2+1)^2}dz \overset{IBP}= \int_0^\infty \frac{a\cos(2az)}{z^2+1}dz \end{align}
Next, let $J(b) = \int_0^\infty \frac{\sin(bz)}{z(z^2+1)}dz $ and
$$J’’(b) = \int_0^\infty \frac{\sin(bz)}{z(z^2+1)}dz - \int_0^\infty \frac{\sin(bz)}zdz =J(b) - \frac\pi2 $$
Solve to get $J(b) = \frac\pi2(1-e^{-b})$ and $I’(a) = a J’(b)|_{b=2a}=\frac\pi2a e^{-2a}$. Then $$I(a)= \int_0^a I’(s)ds = \frac\pi2 \int_0^a se^{-2s}ds= \frac\pi8-\frac\pi8(1+2a)e^{-2a} $$