Residue Applications to Real Integrals

Solution:

Step 1: Poles of $f(z) = \frac{1}{z^2+1} = \frac{1}{(z-i)(z+i)}$ are at $z = \pm i$. Only $z = i$ is in the upper half-plane.

Step 2: Close with semicircle $C_R$ in upper half-plane. On the arc, $|f(z)| \leq \frac{1}{R^2-1}$, so the arc integral vanishes as $R \to \infty$.

Step 3: $\text{Res}(f,i) = \lim_{z\to i} (z-i) \cdot \frac{1}{(z-i)(z+i)} = \frac{1}{2i}$

Step 4: $I = 2\pi i \cdot \frac{1}{2i} = \pi$

Solution:

Substituting $z = e^{i\theta}$: $\cos\theta = \frac{z+z^{-1}}{2}$, $d\theta = \frac{dz}{iz}$

$I = \oint_{|z|=1} \frac{1}{2+\frac{z+z^{-1}}{2}} \cdot \frac{dz}{iz} = \oint_{|z|=1} \frac{2dz}{i(z^2+4z+1)}$

Poles: $z^2+4z+1 = 0$ gives $z = -2 \pm \sqrt{3}$. Only $z_0 = -2+\sqrt{3}$ is inside $|z|=1$.

$\text{Res}(f,z_0) = \lim_{z\to z_0} \frac{2}{i} \cdot \frac{1}{z+2+\sqrt{3}} = \frac{2}{i\cdot 2\sqrt{3}} = \frac{1}{i\sqrt{3}}$

Therefore: $I = 2\pi i \cdot \frac{1}{i\sqrt{3}} = \frac{2\pi}{\sqrt{3}}$

Solution:

Write $I = \text{Re}\left[\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1}dx\right]$

Consider $\int_{-\infty}^{\infty} \frac{e^{iz}}{z^2+1}dz$. Pole at $z=i$ in upper half-plane.

By Jordan's lemma, the arc contribution vanishes. $\text{Res}\left(\frac{e^{iz}}{z^2+1}, i\right) = \frac{e^{i^2}}{2i} = \frac{e^{-1}}{2i}$

Therefore: $\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1}dx = 2\pi i \cdot \frac{e^{-1}}{2i} = \pi e^{-1}$

Taking real part: $I = \frac{\pi}{e}$

Solution:

Since the integrand is even: $I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{x^4+1}$

Poles: $z^4 = -1 = e^{i\pi}$, so $z = e^{i(\pi + 2k\pi)/4}$ for $k = 0,1,2,3$

In the upper half-plane: $z_1 = e^{i\pi/4}$, $z_2 = e^{i3\pi/4}$

Computing residues and applying residue theorem gives $I = \frac{\pi}{2\sqrt{2}}$.

Solution:

Consider $\int_{-\infty}^{\infty} \frac{e^{ix}}{x}dx$ with a small indentation around $z = 0$.

The integral equals $\pi i$ (from the small semicircle contribution). Taking imaginary part: $\int_0^{\infty} \frac{\sin x}{x}dx = \frac{\pi}{2}$.

Solution:

Use a keyhole contour avoiding the branch cut along the positive real axis.

The function $z^{\alpha-1}/(1+z)$ has a simple pole at $z = -1$.

After computing contributions from all parts of the contour and taking the limit, we get:

$\int_0^{\infty} \frac{x^{\alpha-1}}{1+x}dx = \frac{\pi}{\sin(\alpha\pi)}$

Solution:

Poles of $z^2/(z^4+1)$ in upper half-plane: $z = e^{i\pi/4}$, $z = e^{i3\pi/4}$

Computing residues and applying residue theorem:

$\text{Res}(f,e^{i\pi/4}) = \frac{e^{i\pi/2}}{4e^{i3\pi/4}} = \frac{i}{4e^{i3\pi/4}} = \frac{e^{-i3\pi/4}}{4}i$

Similar calculation for the other pole. After summing residues, the integral equals $\frac{\pi}{\sqrt{2}}$.

Solution:

Consider $\int_{-\infty}^{\infty} \frac{e^{iax}}{x^2 + b^2}dx$. The function has a pole at $z = ib$ in the upper half-plane.

$\text{Res}\left(\frac{e^{iaz}}{z^2 + b^2}, ib\right) = \frac{e^{ia(ib)}}{2ib} = \frac{e^{-ab}}{2ib}$

By residue theorem: $\int_{-\infty}^{\infty} \frac{e^{iax}}{x^2 + b^2}dx = 2\pi i \cdot \frac{e^{-ab}}{2ib} = \frac{\pi e^{-ab}}{b}$

Taking real part and using evenness: $\int_0^{\infty} \frac{\cos(ax)}{x^2 + b^2}dx = \frac{\pi e^{-ab}}{2b}$

Solution:

Consider $\int_{-\infty}^{\infty} \frac{xe^{ix}}{x^2 + a^2}dx$. The function has a pole at $z = ia$.

$\text{Res}\left(\frac{ze^{iz}}{z^2 + a^2}, ia\right) = \frac{iae^{i(ia)}}{2ia} = \frac{e^{-a}}{2}$

The integral equals $2\pi i \cdot \frac{e^{-a}}{2} = \pi i e^{-a}$

Taking imaginary part and using oddness: $\int_0^{\infty} \frac{x\sin x}{x^2 + a^2}dx = \frac{\pi e^{-a}}{2}$

Solution:

Using $z = e^{i\theta}$: $\cos\theta = \frac{z+z^{-1}}{2}$, $d\theta = \frac{dz}{iz}$

$I = \oint_{|z|=1} \frac{1}{5 - 3(z+z^{-1})/2} \cdot \frac{dz}{iz} = \oint_{|z|=1} \frac{2dz}{i(10z - 3z^2 - 3)}$

Solving $3z^2 - 10z + 3 = 0$ gives $z = 1/3$ and $z = 3$. Only $z = 1/3$ is inside the unit circle.

Computing the residue and applying residue theorem gives $I = \frac{\pi}{2}$.

Solution:

Since the integrand is even: $I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)(x^2+4)}$

Poles in upper half-plane: $z = i$ and $z = 2i$

Computing residues and summing gives $I = \frac{\pi}{6}$.

Solution:

Using $z = e^{i\theta}$: $I = \oint_{|z|=1} \frac{dz}{iz(3 + z + z^{-1})} = \oint_{|z|=1} \frac{dz}{i(3z + z^2 + 1)}$

Solving $z^2 + 3z + 1 = 0$, only one root is inside the unit circle. Computing the residue gives $I = \frac{2\pi}{\sqrt{5}}$.

Solution:

Factor: $x^4 + 5x^2 + 4 = (x^2+1)(x^2+4)$

Using partial fractions and residue theorem on the upper half-plane gives $I = \frac{\pi}{6}$.

Solution:

Use $\sin^2 x = \frac{1 - \cos 2x}{2}$, so:

$I = \frac{1}{2}\int_0^{\infty} \frac{1 - \cos 2x}{x^2}dx = \frac{1}{2}\int_0^{\infty} \frac{dx}{x^2} - \frac{1}{2}\int_0^{\infty} \frac{\cos 2x}{x^2}dx$

Using integration by parts and known results, $I = \frac{\pi}{2}$.

Solution:

Since the integrand is even: $I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)^{n+1}}$

The function has an $(n+1)$th order pole at $z = i$ in the upper half-plane.

Computing the residue using the derivative formula:

$\text{Res}(f,i) = \frac{1}{n!}\lim_{z\to i} \frac{d^n}{dz^n}\left[\frac{1}{(z+i)^{n+1}}\right]$

After computation: $I = \frac{\pi(2n-1)!!}{2(2n)!!}$

Solution:

Consider $\int_{-\infty}^{\infty} \frac{xe^{ix}}{x^2 + a^2}dx$. The pole at $z = ia$ gives:

$\text{Res}\left(\frac{ze^{iz}}{z^2 + a^2}, ia\right) = \frac{iae^{-a}}{2ia} = \frac{e^{-a}}{2}$

The integral equals $2\pi i \cdot \frac{e^{-a}}{2} = \pi i e^{-a}$

Taking imaginary part: $\int_0^{\infty} \frac{x\sin x}{x^2 + a^2}dx = \frac{\pi e^{-a}}{2}$

Solution:

Using $z = e^{i\theta}$ substitution:

$I = \oint_{|z|=1} \frac{dz}{iz(a + b(z+z^{-1})/2)} = \oint_{|z|=1} \frac{2dz}{i(2az + bz^2 + b)}$

Solving the quadratic and applying residue theorem gives $I = \frac{2\pi}{\sqrt{a^2 - b^2}}$.

Solution:

Using integration by parts and known results for $\int_0^{\infty} \frac{1-\cos(ax)}{x^2}dx$, we get:

$I = \frac{\pi}{2}(q - p)$.

Solution:

Complete the square: $x^2 + 4x + 5 = (x+2)^2 + 1$

The poles are at $z = -2 \pm i$. Only $z = -2 + i$ is in the upper half-plane.

Computing the residue and applying the residue theorem gives $I = \pi e^{-1-2i}$.

Solution:

Using $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$ and known results:

$I = \frac{3\pi}{32}$.

Solution:

Since the integrand is even: $I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{x^4+1}$

The poles in the upper half-plane are at $z = e^{i\pi/4}$ and $z = e^{i3\pi/4}$.

Computing residues and summing gives $I = \frac{\pi}{2\sqrt{2}}$.

Solution:

Using trigonometric identities and integration by parts:

$I = \frac{\pi}{4}(|p+q| - |p-q|)$

Solution:

Write $I = \int_0^{\infty} \frac{1-\cos(bx)}{x^2}dx - \int_0^{\infty} \frac{1-\cos(ax)}{x^2}dx$

Using known results: $\int_0^{\infty} \frac{1-\cos(cx)}{x^2}dx = \frac{\pi c}{2}$

Therefore: $I = \frac{\pi}{2}(b - a)$

Solution:

Using $z = e^{i\theta}$ substitution and computing residues:

$I = \frac{2\pi a}{(a^2-b^2)^{3/2}}$

Solution:

Using partial fractions and residue theorem on the upper half-plane:

$I = \frac{\pi}{6}$

Solution:

On $C_R$, we have $z = Re^{i\theta}$ for $\theta \in [0, \pi]$.

For $z$ in the upper half-plane, $\text{Im}(z) \geq 0$, so $|e^{iz}| = e^{-\text{Im}(z)} \leq 1$.

Also, $|z^2+1| \geq |z|^2 - 1 = R^2 - 1$ for large $R$.

Therefore: $\left|\frac{e^{iz}}{z^2+1}\right| \leq \frac{1}{R^2-1}$

By ML inequality: $\left|\int_{C_R} \frac{e^{iz}}{z^2+1}dz\right| \leq \frac{\pi R}{R^2-1} \to 0$ as $R \to \infty$.

This is a direct application of Jordan's lemma: since $f(z) = 1/(z^2+1) \to 0$ uniformly as $|z| \to \infty$ and $a = 1 > 0$, the integral vanishes.

Solution:

Consider $I = \int_{-\infty}^{\infty} \frac{e^{iax}}{x^2+b^2}dx$, so $\int_0^{\infty} \frac{\cos(ax)}{x^2+b^2}dx = \frac{1}{2}\text{Re}(I)$.

Close the contour in the upper half-plane (since $a > 0$). The poles are at $z = \pm ib$, but only $z = ib$ is in the upper half-plane.

By Jordan's lemma, the integral over the large semicircle vanishes as $R \to \infty$.

Residue at $z = ib$: $\text{Res}\left(\frac{e^{iaz}}{z^2+b^2}, ib\right) = \frac{e^{ia(ib)}}{2ib} = \frac{e^{-ab}}{2ib}$

By residue theorem: $I = 2\pi i \cdot \frac{e^{-ab}}{2ib} = \frac{\pi e^{-ab}}{b}$

Therefore: $\int_0^{\infty} \frac{\cos(ax)}{x^2+b^2}dx = \frac{1}{2} \cdot \frac{\pi e^{-ab}}{b} = \frac{\pi e^{-ab}}{2b}$