Exercises — Residue theorem — computing real integrals

Tool: rational function, denominator degree $2$, numerator degree $0$, so the degree gap is $2\ge 2$ — the semicircle machine in the UHP works (the arc dies like $1/{\rho }^2\cdot \pi \rho \to 0$). Poles: $x^2+9=0$ as a complex equation gives $z^2=-9$, i.e. $z=\pm 3i$. The blue UHP loop encloses only $z=+3i$ (positive imaginary part). The pole $z=-3i$ sits in the lower half — outside our loop, ignored. Answer: UHP semicircle; enclosed pole $z=3i$ only.

Tool: the integrand is a function of $\cos \theta$ over a full period $[0,2\pi ]$ — this is already a loop. Substitute $z=e^{i\theta }$, which traces the unit circle $|z|=1$ once counter-clockwise as $\theta$ runs $0\to 2\pi$. Recall the dictionary: $\cos \theta =\frac{z+z^{-1}}{2}$ and $d\theta =\frac{dz}{iz}$. Answer: substitute $z=e^{i\theta }$; integrate on the orange unit circle $|z|=1$, CCW.

Enclosed pole: $z=3i$ (from L1.1). It is a simple pole (the denominator $z^2+9$ has a single zero there). Residue via $g/h^{\prime }$: with $g=1$, $h=z^2+9$, $h^{\prime }=2z$, ${\mathrm{Res}}_{z=3i}\frac{1}{z^2+9}=\frac{g(3i)}{h^{\prime }(3i)}=\frac{1}{2\cdot 3i}=\frac{1}{6i}.$ Why $g/h^{\prime }$? For a simple pole this avoids factoring — fastest route. Assemble: ${\int }_{-\infty }^{\infty }\frac{dx}{x^2+9}=2\pi i\cdot \frac{1}{6i}=\frac{\pi }{3}.$ Sanity: $\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan \frac{x}{a}$; over $\mathbb{R}$ this is $\frac{\pi }{a}=\frac{\pi }{3}$. ✓

Substitute $z=e^{i\theta }$, $\cos \theta =\frac{z+z^{-1}}{2}$, $d\theta =\frac{dz}{iz}$: ${\oint }_{|z|=1}\frac{1}{5+4\cdot \frac{z+z^{-1}}{2}}\cdot \frac{dz}{iz}={\oint }_{|z|=1}\frac{1}{5+2z+2z^{-1}}\cdot \frac{dz}{iz}.$ Multiply top and bottom by $z$ to clear the $z^{-1}$: $={\oint }_{|z|=1}\frac{1}{iz\left(5+2z+\frac{2}{z}\right)}dz={\oint }_{|z|=1}\frac{1}{i\left(2z^2+5z+2\right)}dz.$ Poles: solve $2z^2+5z+2=0$: $z=\frac{-5\pm \sqrt{25-16}}{4}=\frac{-5\pm 3}{4}$, giving $z=-\frac{1}{2}$ and $z=-2$. Inside $|z|<1$? $|-\frac{1}{2}|=\frac{1}{2}<1$ ✓; $|-2|=2>1$ ✗. Only $z=-\frac{1}{2}$ is trapped. Residue (simple), using $g/h^{\prime }$ with $h=i(2z^2+5z+2)$, $h^{\prime }=i(4z+5)$, $g=1$: ${\mathrm{Res}}_{z=-1/2}=\frac{1}{i(4\cdot (-\frac{1}{2})+5)}=\frac{1}{i(3)}=\frac{1}{3i}.$ Assemble: $2\pi i\cdot \frac{1}{3i}=\frac{2\pi }{3}.$ Sanity: the standard formula ${\int }_0^{2\pi }\frac{d\theta }{a+b\cos \theta }=\frac{2\pi }{\sqrt{a^2-b^2}}$ with $a=5,b=4$ gives $\frac{2\pi }{\sqrt{9}}=\frac{2\pi }{3}$. ✓

Choose the right complex function. We want $\cos (3x)$, but never use $\cos (3z)$ — it grows in the UHP. Instead use $f(z)=\frac{e^{i3z}}{z^2+4},a=3>0,$ because $|e^{i3z}|=e^{-3\mathrm{Im}z}$ decays upward. At the end take the real part. Arc dies by Jordan's Lemma: here $R(z)=\frac{1}{z^2+4}\to 0$ as $|z|\to \infty$; combined with $a=3>0$ the exponential kills the UHP arc even though we'd only get a degree gap of $2$ anyway — Jordan is the clean justification for the oscillatory factor. Enclosed pole: $z^2+4=0\Rightarrow z=\pm 2i$; UHP gives $z=2i$ (simple). Residue via $g/h^{\prime }$, $g=e^{i3z}$, $h^{\prime }=2z$: ${\mathrm{Res}}_{z=2i}=\frac{e^{i3(2i)}}{2(2i)}=\frac{e^{-6}}{4i}.$ Assemble: ${\int }_{-\infty }^{\infty }\frac{e^{i3x}}{x^2+4}dx=2\pi i\cdot \frac{e^{-6}}{4i}=\frac{\pi e^{-6}}{2}.$ Take the real part (the imaginary part is the $\sin$ integral, which is $0$ by oddness): ${\int }_{-\infty }^{\infty }\frac{\cos (3x)}{x^2+4}dx=\frac{\pi }{2e^6}.$

Check the machine applies: numerator degree $2$, denominator degree $4$, gap $=2\ge 2$ ✓, and there are no real poles (roots are $\pm i,\pm 2i$). UHP semicircle, arc dies. Enclosed poles: $z=i$ and $z=2i$ (both simple, both in the UHP). Write $f(z)=\frac{z^2}{(z^2+1)(z^2+4)}$. Use $g/h^{\prime }$ where $h=(z^2+1)(z^2+4)$, so $h^{\prime }(z)=2z(z^2+4)+2z(z^2+1)=2z(2z^2+5).$ At $z=i$ ($z^2=-1$): $g=z^2=-1$, $h^{\prime }=2i(2(-1)+5)=2i(3)=6i$, so ${\mathrm{Res}}_i=\frac{-1}{6i}.$ At $z=2i$ ($z^2=-4$): $g=z^2=-4$, $h^{\prime }=2(2i)(2(-4)+5)=4i(-3)=-12i$, so ${\mathrm{Res}}_{2i}=\frac{-4}{-12i}=\frac{1}{3i}.$ Sum: $\frac{-1}{6i}+\frac{1}{3i}=\frac{-1+2}{6i}=\frac{1}{6i}.$ Assemble: $2\pi i\cdot \frac{1}{6i}=\frac{\pi }{3}.$ Answer: ${\int }_{-\infty }^{\infty }\frac{x^2}{(x^2+1)(x^2+4)}dx=\frac{\pi }{3}.$

Set-up. We have $\sin x$, so use $f(z)=\frac{z{e}^{iz}}{z^2+1}$ (with $a=1$) and take the imaginary part at the end. Arc dies by Jordan's Lemma: $R(z)=\frac{z}{z^2+1}\to 0$ as $|z|\to \infty$ (numerator degree $1$, denominator $2$), and $a=1>0$, so the UHP arc vanishes — note the raw degree gap is only $1$, so Jordan is essential here; the plain rational bound would fail. Enclosed pole: $z=i$ (simple). Residue with $g=z{e}^{iz}$, $h^{\prime }=2z$: ${\mathrm{Res}}_i=\frac{i{e}^{i\cdot i}}{2i}=\frac{e^{-1}}{2}.$ Assemble: ${\int }_{-\infty }^{\infty }\frac{x{e}^{ix}}{x^2+1}dx=2\pi i\cdot \frac{e^{-1}}{2}=\frac{\pi i}{e}.$ Take the imaginary part (this is $\int \frac{x\sin x}{x^2+1}dx$; the real part is $\int \frac{x\cos x}{x^2+1}dx=0$ by oddness): ${\int }_{-\infty }^{\infty }\frac{x\sin x}{x^2+1}dx=\frac{\pi }{e}.$

Why a principal value? The integrand $\frac{\sin x}{x}$ is actually finite at $x=0$ (limit $1$), but our complex tool $f(z)=\frac{e^{iz}}{z}$ has a genuine simple pole at $z=0$ sitting on the real axis — right on the contour. We handle it with an indented contour: a tiny semicircle of radius $\epsilon$ that detours over the pole, plus a big UHP semicircle. See Principal Value Integrals.

The closed contour (real axis with a small bump around $0$, closed by the big arc) encloses no poles, so by the residue theorem the whole loop integral is $0$: $\underset{\text{real parts}}{\underbrace{{\int }_{-\rho }^{-\epsilon }+{\int }_{\epsilon }^{\rho }}}+\underset{\text{small arc}}{\underbrace{{\int }_{{\gamma }_{\epsilon }}}}+\underset{\text{big arc}}{\underbrace{{\int }_{{\Gamma }_{\rho }}}}=0.$ Big arc ${\Gamma }_{\rho }\to 0$ by Jordan's Lemma ($R=1/z\to 0$, $a=1>0$). Small arc ${\gamma }_{\epsilon }$: a clockwise semicircle around a simple pole contributes $-i\pi \cdot {\mathrm{Res}}_{z=0}$ (half of $2\pi i$, minus for clockwise). Here ${\mathrm{Res}}_0\frac{e^{iz}}{z}=e^0=1$, so ${\int }_{{\gamma }_{\epsilon }}\to -i\pi$. Let $\rho \to \infty ,\epsilon \to 0$: $P.V.{\int }_{-\infty }^{\infty }\frac{e^{ix}}{x}dx+(-i\pi )=0\Rightarrow P.V.{\int }_{-\infty }^{\infty }\frac{e^{ix}}{x}dx=i\pi .$ Take the imaginary part ($\frac{\sin x}{x}$ is the imaginary part of $\frac{e^{ix}}{x}$; the real part $\frac{\cos x}{x}$ is odd and gives $0$ in P.V.): ${\int }_{-\infty }^{\infty }\frac{\sin x}{x}dx=\pi .$

Machine check: denominator degree $4$, numerator degree $0$, gap $2$ — UHP semicircle, arc dies. Enclosed pole: $(z^2+a^2{)}^2=(z-ai{)}^2(z+ai{)}^2$, so $z=ai$ is a pole of order $m=2$ in the UHP. Order-2 residue (differentiate $m-1=1$ time): ${\mathrm{Res}}_{ai}=\frac{1}{1!}\frac{d}{dz}{\left[(z-ai{)}^2\frac{1}{(z-ai{)}^2(z+ai{)}^2}\right]}_{z=ai}=\frac{d}{dz}(z+ai{)}^{-2}{|}_{ai}.$ $\frac{d}{dz}(z+ai{)}^{-2}=\frac{-2}{(z+ai{)}^3}\Rightarrow \text{at }z=ai:\frac{-2}{(2ai{)}^3}=\frac{-2}{8a^3{i}^3}=\frac{-2}{-8a^{3}i}=\frac{1}{4a^{3}i}.$ Assemble: ${\int }_{-\infty }^{\infty }\frac{dx}{(x^2+a^2{)}^2}=2\pi i\cdot \frac{1}{4a^{3}i}=\frac{\pi }{2a^3}.$ Check $a=1$: $\frac{\pi }{2}$ ✓ — matches the parent note's double-pole example.

Substitute $z=e^{i\theta }$, $\cos \theta =\frac{z+z^{-1}}{2}=\frac{z^2+1}{2z}$, $d\theta =\frac{dz}{iz}$. Then $2+\cos \theta =\frac{4z+z^2+1}{2z}=\frac{z^2+4z+1}{2z}.$ So the integrand becomes $\frac{1}{(2+\cos \theta {)}^2}d\theta =\frac{(2z{)}^2}{(z^2+4z+1{)}^2}\cdot \frac{dz}{iz}=\frac{4z^2}{iz(z^2+4z+1{)}^2}dz=\frac{4z}{i(z^2+4z+1{)}^2}dz.$ Poles: $z^2+4z+1=0\Rightarrow z=-2\pm \sqrt{3}$. Inside $|z|<1$: $z_0=-2+\sqrt{3}\approx -0.268$ ✓ (order $2$); $z_1=-2-\sqrt{3}\approx -3.73$ ✗. The extra $\frac{1}{iz}$ did not add a pole at $z=0$ because the numerator $4z$ cancels it — good, check confirmed. Write $g(z)=\frac{4z}{i(z-z_1{)}^2}$ so that $f=\frac{g(z)}{(z-z_0{)}^2}$ near $z_0$; the order-2 residue is ${\mathrm{Res}}_{z_0}=\frac{d}{dz}[(z-z_0{)}^{2}f{]}_{z_0}=g^{\prime }(z_0).$ Let $c=z_0-z_1=(-2+\sqrt{3})-(-2-\sqrt{3})=2\sqrt{3}$. With $g(z)=\frac{4z}{i(z-z_1{)}^2}$, $g^{\prime }(z)=\frac{4}{i}\cdot \frac{(z-z_1{)}^2-z\cdot 2(z-z_1)}{(z-z_1{)}^4}=\frac{4}{i}\cdot \frac{(z-z_1)-2z}{(z-z_1{)}^3}=\frac{4}{i}\cdot \frac{-z-z_1}{(z-z_1{)}^3}.$ At $z=z_0$: $z-z_1=c=2\sqrt{3}$, $-z_0-z_1=-(z_0+z_1)$. Since $z_0+z_1=-4$ (sum of roots), $-z_0-z_1=4$. Thus $g^{\prime }(z_0)=\frac{4}{i}\cdot \frac{4}{(2\sqrt{3}{)}^3}=\frac{16}{i\cdot 24\sqrt{3}}=\frac{2}{3\sqrt{3}i}.$ Assemble: $2\pi i\cdot \frac{2}{3\sqrt{3}i}=\frac{4\pi }{3\sqrt{3}}.$ Sanity: the known formula ${\int }_0^{2\pi }\frac{d\theta }{(a+\cos \theta {)}^2}=\frac{2\pi a}{(a^2-1{)}^{3/2}}$ with $a=2$ gives $\frac{4\pi }{3^{3/2}}=\frac{4\pi }{3\sqrt{3}}$ ✓. Answer: ${\int }_0^{2\pi }\frac{d\theta }{(2+\cos \theta {)}^2}=\frac{4\pi }{3\sqrt{3}}=\frac{4\sqrt{3}\pi }{9}.$

Pick the loop ::: UHP semicircle for $(-\infty ,\infty )$, unit circle for $[0,2\pi ]$; indent around real-axis poles for P.V. Kill the arc ::: degree gap $\ge 2$ (rational) or Jordan's Lemma (oscillatory $e^{iaz}$, $a>0$). Finish ::: sum residues of enclosed poles only, multiply by $2\pi i$; a clockwise half-indent gives $-i\pi \mathrm{Res}$.