Exercises — Residue theorem — computing real integrals

Tool: rational function, denominator degree $2$, numerator degree $0$, toh degree gap $2\ge 2$ hai — UHP mein semicircle machine kaam karti hai (arc $1/{\rho }^2\cdot \pi \rho \to 0$ ki tarah die hoti hai). Poles: $x^2+9=0$ ko complex equation ki tarah solve karo: $z^2=-9$, yani $z=\pm 3i$. Blue UHP loop sirf $z=+3i$ ko enclose karti hai (positive imaginary part). Pole $z=-3i$ lower half mein baithta hai — hamare loop ke bahar, ignored. Answer: UHP semicircle; enclosed pole sirf $z=3i$.

Tool: integrand ek full period $[0,2\pi ]$ par $\cos \theta$ ka function hai — yeh already ek loop hai. $z=e^{i\theta }$ substitute karo, jo unit circle $|z|=1$ ko counter-clockwise ek baar trace karta hai jab $\theta$ $0\to 2\pi$ tak jaata hai. Dictionary yaad karo: $\cos \theta =\frac{z+z^{-1}}{2}$ aur $d\theta =\frac{dz}{iz}$. Answer: $z=e^{i\theta }$ substitute karo; orange unit circle $|z|=1$ par CCW integrate karo.

Enclosed pole: $z=3i$ (L1.1 se). Yeh ek simple pole hai (denominator $z^2+9$ wahan ek single zero rakhta hai). Residue via $g/h^{\prime }$: $g=1$, $h=z^2+9$, $h^{\prime }=2z$ ke saath, ${\mathrm{Res}}_{z=3i}\frac{1}{z^2+9}=\frac{g(3i)}{h^{\prime }(3i)}=\frac{1}{2\cdot 3i}=\frac{1}{6i}.$ $g/h^{\prime }$ kyun? Simple pole ke liye yeh factoring se bachata hai — sabse fast route. Assemble: ${\int }_{-\infty }^{\infty }\frac{dx}{x^2+9}=2\pi i\cdot \frac{1}{6i}=\frac{\pi }{3}.$ Sanity: $\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan \frac{x}{a}$; $\mathbb{R}$ par yeh $\frac{\pi }{a}=\frac{\pi }{3}$ hai. ✓

Substitute $z=e^{i\theta }$, $\cos \theta =\frac{z+z^{-1}}{2}$, $d\theta =\frac{dz}{iz}$: ${\oint }_{|z|=1}\frac{1}{5+4\cdot \frac{z+z^{-1}}{2}}\cdot \frac{dz}{iz}={\oint }_{|z|=1}\frac{1}{5+2z+2z^{-1}}\cdot \frac{dz}{iz}.$ $z^{-1}$ clear karne ke liye upar aur neeche $z$ se multiply karo: $={\oint }_{|z|=1}\frac{1}{iz\left(5+2z+\frac{2}{z}\right)}dz={\oint }_{|z|=1}\frac{1}{i\left(2z^2+5z+2\right)}dz.$ Poles: $2z^2+5z+2=0$ solve karo: $z=\frac{-5\pm \sqrt{25-16}}{4}=\frac{-5\pm 3}{4}$, jo $z=-\frac{1}{2}$ aur $z=-2$ deta hai. $|z|<1$ ke andar? $|-\frac{1}{2}|=\frac{1}{2}<1$ ✓; $|-2|=2>1$ ✗. Sirf $z=-\frac{1}{2}$ trapped hai. Residue (simple), $g/h^{\prime }$ use karke $h=i(2z^2+5z+2)$, $h^{\prime }=i(4z+5)$, $g=1$ ke saath: ${\mathrm{Res}}_{z=-1/2}=\frac{1}{i(4\cdot (-\frac{1}{2})+5)}=\frac{1}{i(3)}=\frac{1}{3i}.$ Assemble: $2\pi i\cdot \frac{1}{3i}=\frac{2\pi }{3}.$ Sanity: standard formula ${\int }_0^{2\pi }\frac{d\theta }{a+b\cos \theta }=\frac{2\pi }{\sqrt{a^2-b^2}}$ mein $a=5,b=4$ dalne par $\frac{2\pi }{\sqrt{9}}=\frac{2\pi }{3}$ milta hai. ✓

Sahi complex function choose karo. Hum $\cos (3x)$ chahte hain, lekin kabhi bhi $\cos (3z)$ use mat karo — yeh UHP mein grow karta hai. Balki use karo $f(z)=\frac{e^{i3z}}{z^2+4},a=3>0,$ kyunki $|e^{i3z}|=e^{-3\mathrm{Im}z}$ upar ki taraf decay karta hai. Ant mein real part lo. Arc Jordan's Lemma se die hoti hai: yahan $R(z)=\frac{1}{z^2+4}\to 0$ jab $|z|\to \infty$; $a=3>0$ ke saath exponential UHP arc ko kill kar deta hai chahe degree gap waise bhi $2$ hi hota — oscillatory factor ke liye Jordan clean justification hai. Enclosed pole: $z^2+4=0\Rightarrow z=\pm 2i$; UHP se $z=2i$ milta hai (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: ${\int }_{-\infty }^{\infty }\frac{e^{i3x}}{x^2+4}dx=2\pi i\cdot \frac{e^{-6}}{4i}=\frac{\pi e^{-6}}{2}.$ Real part lo (imaginary part $\sin$ integral hai, jo oddness se $0$ hai): ${\int }_{-\infty }^{\infty }\frac{\cos (3x)}{x^2+4}dx=\frac{\pi }{2e^6}.$

Check karo ki machine apply hoti hai: numerator degree $2$, denominator degree $4$, gap $=2\ge 2$ ✓, aur koi real poles nahi hain (roots $\pm i,\pm 2i$ hain). UHP semicircle, arc die hoti hai. Enclosed poles: $z=i$ aur $z=2i$ (dono simple, dono UHP mein). $f(z)=\frac{z^2}{(z^2+1)(z^2+4)}$ likho. $g/h^{\prime }$ use karo jahan $h=(z^2+1)(z^2+4)$, toh $h^{\prime }(z)=2z(z^2+4)+2z(z^2+1)=2z(2z^2+5).$ $z=i$ par ($z^2=-1$): $g=z^2=-1$, $h^{\prime }=2i(2(-1)+5)=2i(3)=6i$, toh ${\mathrm{Res}}_i=\frac{-1}{6i}.$ $z=2i$ par ($z^2=-4$): $g=z^2=-4$, $h^{\prime }=2(2i)(2(-4)+5)=4i(-3)=-12i$, toh ${\mathrm{Res}}_{2i}=\frac{-4}{-12i}=\frac{1}{3i}.$ 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. Hamare paas $\sin x$ hai, toh $f(z)=\frac{z{e}^{iz}}{z^2+1}$ use karo ($a=1$ ke saath) aur ant mein imaginary part lo. Arc Jordan's Lemma se die hoti hai: $R(z)=\frac{z}{z^2+1}\to 0$ jab $|z|\to \infty$ (numerator degree $1$, denominator $2$), aur $a=1>0$, toh UHP arc vanish hoti hai — note karo ki raw degree gap sirf $1$ hai, toh Jordan yahan essential hai; plain rational bound fail ho jaata. Enclosed pole: $z=i$ (simple). Residue $g=z{e}^{iz}$, $h^{\prime }=2z$ ke saath: ${\mathrm{Res}}_i=\frac{i{e}^{i\cdot i}}{2i}=\frac{e^{-1}}{2}.$ Assemble: ${\int }_{-\infty }^{\infty }\frac{x{e}^{ix}}{x^2+1}dx=2\pi i\cdot \frac{e^{-1}}{2}=\frac{\pi i}{e}.$ Imaginary part lo (yeh $\int \frac{x\sin x}{x^2+1}dx$ hai; real part $\int \frac{x\cos x}{x^2+1}dx=0$ oddness se hai): ${\int }_{-\infty }^{\infty }\frac{x\sin x}{x^2+1}dx=\frac{\pi }{e}.$

Principal value kyun? Integrand $\frac{\sin x}{x}$ actually $x=0$ par finite hai (limit $1$ hai), lekin humara complex tool $f(z)=\frac{e^{iz}}{z}$ $z=0$ par genuinely ek simple pole rakhta hai jo real axis par baitha hai — bilkul contour par. Hum isse indented contour se handle karte hain: radius $\epsilon$ ka ek tiny semicircle jo pole ke upar detour karta hai, plus ek bada UHP semicircle. Dekho Principal Value Integrals.

Closed contour ($0$ ke aas-paas ek small bump ke saath real axis, bade arc se closed) koi poles enclose nahi karta, toh residue theorem se poora loop integral $0$ hai: $\underset{\text{real parts}}{\underbrace{{\int }_{-\rho }^{-\epsilon }+{\int }_{\epsilon }^{\rho }}}+\underset{\text{small arc}}{\underbrace{{\int }_{{\gamma }_{\epsilon }}}}+\underset{\text{big arc}}{\underbrace{{\int }_{{\Gamma }_{\rho }}}}=0.$ Big arc ${\Gamma }_{\rho }\to 0$ Jordan's Lemma se ($R=1/z\to 0$, $a=1>0$). Small arc ${\gamma }_{\epsilon }$: ek simple pole ke around clockwise semicircle $-i\pi \cdot {\mathrm{Res}}_{z=0}$ contribute karti hai ($2\pi i$ ka half, clockwise ke liye minus). Yahan ${\mathrm{Res}}_0\frac{e^{iz}}{z}=e^0=1$, toh ${\int }_{{\gamma }_{\epsilon }}\to -i\pi$. $\rho \to \infty ,\epsilon \to 0$ lene par: $P.V.{\int }_{-\infty }^{\infty }\frac{e^{ix}}{x}dx+(-i\pi )=0\Rightarrow P.V.{\int }_{-\infty }^{\infty }\frac{e^{ix}}{x}dx=i\pi .$ Imaginary part lo ($\frac{\sin x}{x}$ $\frac{e^{ix}}{x}$ ka imaginary part hai; real part $\frac{\cos x}{x}$ odd hai aur P.V. mein $0$ deta hai): ${\int }_{-\infty }^{\infty }\frac{\sin x}{x}dx=\pi .$

Machine check: denominator degree $4$, numerator degree $0$, gap $2$ — UHP semicircle, arc die hoti hai. Enclosed pole: $(z^2+a^2{)}^2=(z-ai{)}^2(z+ai{)}^2$, toh $z=ai$ UHP mein order $m=2$ ka pole hai. Order-2 residue ($m-1=1$ baar differentiate karo): ${\mathrm{Res}}_{ai}=\frac{1}{1!}\frac{d}{dz}{\left[(z-ai{)}^2\frac{1}{(z-ai{)}^2(z+ai{)}^2}\right]}_{z=ai}=\frac{d}{dz}(z+ai{)}^{-2}{|}_{ai}.$ $\frac{d}{dz}(z+ai{)}^{-2}=\frac{-2}{(z+ai{)}^3}\Rightarrow \text{at }z=ai:\frac{-2}{(2ai{)}^3}=\frac{-2}{8a^3{i}^3}=\frac{-2}{-8a^{3}i}=\frac{1}{4a^{3}i}.$ Assemble: ${\int }_{-\infty }^{\infty }\frac{dx}{(x^2+a^2{)}^2}=2\pi i\cdot \frac{1}{4a^{3}i}=\frac{\pi }{2a^3}.$ $a=1$ check: $\frac{\pi }{2}$ ✓ — parent note ke double-pole example se match karta hai.

Substitute $z=e^{i\theta }$, $\cos \theta =\frac{z+z^{-1}}{2}=\frac{z^2+1}{2z}$, $d\theta =\frac{dz}{iz}$. Tab $2+\cos \theta =\frac{4z+z^2+1}{2z}=\frac{z^2+4z+1}{2z}.$ Toh integrand ban jaata hai $\frac{1}{(2+\cos \theta {)}^2}d\theta =\frac{(2z{)}^2}{(z^2+4z+1{)}^2}\cdot \frac{dz}{iz}=\frac{4z^2}{iz(z^2+4z+1{)}^2}dz=\frac{4z}{i(z^2+4z+1{)}^2}dz.$ Poles: $z^2+4z+1=0\Rightarrow z=-2\pm \sqrt{3}$. $|z|<1$ ke andar: $z_0=-2+\sqrt{3}\approx -0.268$ ✓ (order $2$); $z_1=-2-\sqrt{3}\approx -3.73$ ✗. Extra $\frac{1}{iz}$ ne $z=0$ par pole nahi add kiya kyunki numerator $4z$ ise cancel kar deta hai — achha, check confirmed. $g(z)=\frac{4z}{i(z-z_1{)}^2}$ likho taaki $f=\frac{g(z)}{(z-z_0{)}^2}$ $z_0$ ke paas ho; order-2 residue hai ${\mathrm{Res}}_{z_0}=\frac{d}{dz}[(z-z_0{)}^{2}f{]}_{z_0}=g^{\prime }(z_0).$ $c=z_0-z_1=(-2+\sqrt{3})-(-2-\sqrt{3})=2\sqrt{3}$ lo. $g(z)=\frac{4z}{i(z-z_1{)}^2}$ ke saath, $g^{\prime }(z)=\frac{4}{i}\cdot \frac{(z-z_1{)}^2-z\cdot 2(z-z_1)}{(z-z_1{)}^4}=\frac{4}{i}\cdot \frac{(z-z_1)-2z}{(z-z_1{)}^3}=\frac{4}{i}\cdot \frac{-z-z_1}{(z-z_1{)}^3}.$ $z=z_0$ par: $z-z_1=c=2\sqrt{3}$, $-z_0-z_1=-(z_0+z_1)$. Kyunki $z_0+z_1=-4$ (roots ka sum), $-z_0-z_1=4$. Toh $g^{\prime }(z_0)=\frac{4}{i}\cdot \frac{4}{(2\sqrt{3}{)}^3}=\frac{16}{i\cdot 24\sqrt{3}}=\frac{2}{3\sqrt{3}i}.$ Assemble: $2\pi i\cdot \frac{2}{3\sqrt{3}i}=\frac{4\pi }{3\sqrt{3}}.$ Sanity: known formula ${\int }_0^{2\pi }\frac{d\theta }{(a+\cos \theta {)}^2}=\frac{2\pi a}{(a^2-1{)}^{3/2}}$ mein $a=2$ dalne par $\frac{4\pi }{3^{3/2}}=\frac{4\pi }{3\sqrt{3}}$ milta hai ✓. Answer: ${\int }_0^{2\pi }\frac{d\theta }{(2+\cos \theta {)}^2}=\frac{4\pi }{3\sqrt{3}}=\frac{4\sqrt{3}\pi }{9}.$

Loop chuno ::: UHP semicircle $(-\infty ,\infty )$ ke liye, unit circle $[0,2\pi ]$ ke liye; real-axis poles ke liye P.V. mein indent karo. Arc ko kill karo ::: degree gap $\ge 2$ (rational) ya Jordan's Lemma (oscillatory $e^{iaz}$, $a>0$). Finish karo ::: sirf enclosed poles ke residues ka sum karo, $2\pi i$ se multiply karo; clockwise half-indent $-i\pi \mathrm{Res}$ deta hai.