The Residue Theorem and the Evaluation of Real Integrals — Examples

The only pole of $R$ in the upper half-plane is $i$, and it is simple with $\mathrm{Res}(R,i)={\lim }_{z\to i}\frac{z-i}{(z-i)(z+i)}=\frac{1}{2i}\mathrm{.}$

Applying [L1] gives ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^2}=2\pi i\cdot \frac{1}{2i}=\pi \mathrm{.}$

The poles of $R$ in the upper half-plane are $a_1=e^{i\pi \mathrm{/}4}$ and $a_2=e^{3i\pi \mathrm{/}4}$, and they are simple. Since $(z^4+1{)}^{\mathrm{\prime }}=4z^3$, $\mathrm{Res}(R,a_j)=\frac{1}{4a_j^3}(j=1,2)\mathrm{.}$

Their sum is $\frac{1}{4}(e^{-3i\pi \mathrm{/}4}+e^{-9i\pi \mathrm{/}4})=\frac{1}{4}(e^{-3i\pi \mathrm{/}4}+e^{-i\pi \mathrm{/}4})=-\frac{i}{2\sqrt{2}}\mathrm{.}$

Therefore [L1] gives ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^4}=2\pi i(-\frac{i}{2\sqrt{2}})=\frac{\pi }{\sqrt{2}}\mathrm{.}$

Apply [L1] with $\lambda =1$. The only upper-half-plane pole is $i$, and $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{e^{iz}}{1+z^2},i)=\frac{e^{ii}}{2i}=\frac{e^{-1}}{2i}\mathrm{.}$

Hence ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{1+x^2}dx=2\pi i\cdot \frac{e^{-1}}{2i}=\frac{\pi }{e}\mathrm{.}$ The value is real, so taking real parts yields ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos x}{1+x^2}dx=\frac{\pi }{e}\mathrm{.}$

On the upper semicircle $z=T{e}^{it}$ one has $\mathrm{\mid }{e}^{iz}\mathrm{\mid }=e^{-T\sin t}$. Convexity of $-\sin t$ on $[0,\pi \mathrm{/}2]$ gives $\sin t\ge 2t\mathrm{/}\pi$ there, and symmetry gives the corresponding bound on the other half. Hence $\mid {\int }_{{\gamma }_T^{+}}\frac{e^{iz}}{z}dz\mid \le {\int }_0^{\pi }{e}^{-T\sin t}dt\le \frac{\pi }{T},$ so the outer arc tends to $0$.

Apply [L1] to the upper contour of radius $T$ with an upper indentation of radius $\epsilon$ at $0$. The contour encloses no pole, while the residue of $e^{iz}\mathrm{/}z$ at $0$ is $1$. Therefore the sum of the two punctured straight integrals, the indentation, and the outer arc is $0$. Letting $\epsilon \downarrow 0$ in this coupled symmetric truncation, then $T\to \mathrm{\infty }$, [L2] and step 1.1 give ${\lim }_{T\to \mathrm{\infty }}{\lim }_{\epsilon \downarrow 0}({\int }_{-T}^{-\epsilon }\frac{e^{ix}}{x}dx+{\int }_{\epsilon }^T\frac{e^{ix}}{x}dx)=i\pi \mathrm{.}$

The imaginary integrand $\sin x\mathrm{/}x$ has the removable value $1$ at $0$ and is locally integrable on the real line, so the imaginary part of step 2.1 is exactly the whole-line principal value in [L3]. Hence $\mathrm{PV}\text{\hspace{0.17em}⁣}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin x}{x}dx=\pi \mathrm{.}$ Since $\sin x\mathrm{/}x$ is even, the symmetric principal value is twice the half-line integral, so ${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=\frac{\pi }{2}$.

By [L1], ${\int }_0^{2\pi }\frac{d\theta }{a+\cos \theta }={\int }_{\mathrm{\mid }z\mathrm{\mid }=1}\frac{2dz}{i(z^2+2az+1)}\mathrm{.}$ The quadratic factors as $z^2+2az+1=(z-z_{+})(z-z_{-}),z_{\pm }=-a\pm \sqrt{a^2-1}\mathrm{.}$

Since $z_{+}{z}_{-}=1$, exactly one root lies inside the unit circle. It is $z_{+}=-a+\mathrm{sgn}(a)\sqrt{a^2-1}$. The residue there is $\frac{2}{i(z_{+}-z_{-})}=\frac{\mathrm{sgn}(a)}{i\sqrt{a^2-1}}\mathrm{.}$ Therefore ${\int }_0^{2\pi }\frac{d\theta }{a+\cos \theta }=2\pi i\cdot \frac{\mathrm{sgn}(a)}{i\sqrt{a^2-1}}=\frac{2\pi \mathrm{sgn}(a)}{\sqrt{a^2-1}}\mathrm{.}$

The factor $R(z)=1\mathrm{/}(1+z)$ has no pole on $[0,\mathrm{\infty })$. Since $0<\mathrm{Re}\alpha <1$, the absolute value of the real integrand is $O(x^{\mathrm{Re}\alpha -1})$ near $0$ and $O(x^{\mathrm{Re}\alpha -2})$ at infinity, so the improper integral converges. On the inner keyhole circle the arc integral is $O({\epsilon }^{\mathrm{Re}\alpha })$, and on the outer circle it is $O(R^{\mathrm{Re}\alpha -1})$; both tend to $0$. Thus every hypothesis of [L1] holds.

The only pole away from the positive real axis is the simple pole at $-1$. On the chosen branch, $(-1{)}^{\alpha -1}=e^{i\pi (\alpha -1)}=-e^{i\pi \alpha }$, so $\mathrm{Res}(f,-1)=-e^{i\pi \alpha }$.

Applying [L1] using step 1.1 and substituting the residue from step 1.2 gives $(1-e^{2\pi i\alpha }){\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{1+x}dx=-2\pi i{e}^{i\pi \alpha }\mathrm{.}$ Since $1-e^{2\pi i\alpha }=-2i{e}^{i\pi \alpha }\sin (\pi \alpha )$, division yields ${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{1+x}dx=\frac{\pi }{\sin (\pi \alpha )}$.

Integrate $F$ around the rectangle with vertices [given, L1] $-T,T,T+ib,-T+ib$. Since $F$ is entire, makes the total integral $0$. The vertical sides vanish as $T\to \mathrm{\infty }$ because $\mathrm{\mid }F(x+iy)\mathrm{\mid }=e^{-x^2+y^2-2by}$ and therefore decays like $e^{-T^2}$ there.

The top horizontal side is traversed from $T+ib$ to $-T+ib$ and therefore contributes $-e^{-b^2}{\int }_{-T}^T{e}^{-x^2}dx\mathrm{.}$ Hence step 1.1 gives ${\int }_{-T}^T{e}^{-x^2}{e}^{2ibx}dx=e^{-b^2}{\int }_{-T}^T{e}^{-x^2}dx+o(1)\mathrm{.}$

Letting $T\to \mathrm{\infty }$ and using [L2] yields ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}{e}^{2ibx}dx=e^{-b^2}\sqrt{\pi }\mathrm{.}$ Taking real parts and then using the evenness of $e^{-x^2}\cos (2bx)$ gives the half-line formula.

The corollary already proves that $2{\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{3}\mathrm{.}$

Dividing by $2$ yields the usual one-sided Basel sum ${\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}\mathrm{.}$

By, the oscillatory integral of $\sin x\mathrm{/}x$ converges on $[1,\mathrm{\infty })$, [L1] so the half-line sine integral is conditionally convergent.

On each interval [L2, algebra] $I_k=[k\pi +\pi \mathrm{/}6,k\pi +5\pi \mathrm{/}6]$ one has $\mathrm{\mid }\sin x\mathrm{\mid }\ge 1\mathrm{/}2$, while $x\le (k+1)\pi$. Therefore ${\int }_{I_k}\mid \frac{\sin x}{x}\mid dx\ge \frac{1}{2}\cdot \frac{2\pi \mathrm{/}3}{(k+1)\pi }=\frac{1}{3(k+1)}\mathrm{.}$ The lower bounds have divergent harmonic sum, so implies ${\int }_1^{\mathrm{\infty }}\mathrm{\mid }\sin x\mathrm{\mid }\mathrm{/}xdx=\mathrm{\infty }$.

Thus $\sin x\mathrm{/}x$ gives a convergent improper integral whose absolute-value integral diverges.

On the upper semicircle ${\gamma }_R^{+}(t)=R{e}^{it}$ one has ${\int }_{{\gamma }_R^{+}}\frac{dz}{z}={\int }_0^{\pi }\frac{iR{e}^{it}}{R{e}^{it}}dt=i\pi \mathrm{.}$

So the arc integral is constant, not vanishing, even though $1\mathrm{/}(R{e}^{it})$ is pointwise $O(R^{-1})$ on the arc. This is the concrete witness behind FALSE: pointwise decay alone makes every large semicircle integral vanish.

The point $a=1$ is a pole of order $2$ of $f$, and [L2] gives the correct residue $\mathrm{Res}(f,1)=\frac{d}{dz}(e^z){\mid }_{z=1}=e\mathrm{.}$

If one incorrectly applies the simple-pole rule from [L1], one obtains ${\lim }_{z\to 1}(z-1)\frac{e^z}{(z-1{)}^2}={\lim }_{z\to 1}\frac{e^z}{z-1},$ which does not exist as a finite complex number. So the simple-pole rule does not recover the residue at a double pole.