$\newcommand{\abs}[1]{\left|#1\right|}
\newcommand{\min}{\mmlToken{mi}{min}}
\newcommand{\d}{\mathrm{~d}}
\DeclareMathOperator{\Re}{Re}
\DeclareMathOperator{\Im}{Im}
\newcommand{\res}[1]{\operatorname*{Res}_{#1}}$
- Prove, for $a>0$, that
\[
\int_{-∞}^∞\frac{\d x}{x^4+a^4}=\fracπ{a^3 \sqrt2}
\]
Solution.
Consider the sector $C$ of angle $\frac{2\pi}{n}$ and an edge $[0,R]$. By Jordan's lemma, the integral along the arc tends to 0 as $R→∞$. So
$$\lim_{R→∞}\int_C \frac{\d z}{1+z^n} = \int_0^∞\frac{\d x}{1+x^n} - \int_0^∞\frac{e^{2\pi i/n} dx}{1+(e^{2\pi i/n} x)^n}=\left(1-e^{2\pi i/n}\right) \int_0^∞\frac{\d x}{1+x^n}$$
Inside $C$, the integrand $\frac1{1+z^n}$ has one simple pole at $z=e^{\pi i/n}$.
\[\res{z=e^{2\pi i/n}}\frac1{1+z^n}=\left.\frac1{nz^{n-1}}\right|_{z=e^{2\pi i/n}}=\frac1{ne^{2\pi i(n-1)/n}}\]
By Residue Theorem,
$$\left ( 1-e^{2\pi i/n}\right) \int_0^∞ \frac{\d x}{1+x^n} = \frac{2 \pi i}{n e^{i \pi (n-1)/n}}$$
The final result is
$$\int_0^∞ \frac{\d x}{1+x^n} = \frac{\pi}{n \sin{(\pi/n)}}$$
Substituting $n=4,x=\frac ya$, we get $\int_{-∞}^∞\frac{\d y}{y^4+a^4}=2\int_0^∞\frac{\d y}{y^4+a^4}=\fracπ{a^3\sqrt2}$
- Show that$$\int_{-∞}^∞\frac{\cos πx}{2x-1}\d x=-\frac π2$$
Solution.
$f(z)=\frac{e^{iπz}}{2z-1}$ has a simple pole at $z=\frac12$. Consider the red contour in the upper half plane.
$$\res{z=\frac12}f(z)=\lim_{z→\frac12}\left(z-\frac12\right)\frac{e^{iπz}}{2z-1}=\left.\frac{e^{iπz}}2\right|_{z=\frac12}=\frac i2$$By Lemma 11.11. the integral of $f$ along the clockwise arc with center $\frac12$ and radius $ϵ→0$ is $\frac i2⋅(0-π)i=\fracπ2$.
$\lim_{z→∞}\frac1{2z-1}=0$, by Jordan's lemma, the integral of $f$ along the anticlockwise arc with radius $R→∞$ is 0.
Because there are no poles inside the contour, the integral is zero, we have $\fracπ2+\int_{-∞}^∞f(z)\d z=0$, so $\int_{-∞}^∞f(z)\d z=-\fracπ2$.
Taking real parts, we get $\int_{-∞}^∞\frac{\cos{\pi x}}{2 x-1}\mathrm{d}x= -\fracπ2$. - Write down a definition of a branch of $\log(z+i)$ which is holomorphic in the cut-plane region.$$ℂ∖\{z:\Re z=0,\Im z ≤-1\}$$
By integrating $\log(z+i)/\left(z^2+1\right)$ around a suitable closed path, evaluate
\[
\int_{-∞}^∞\frac{\log(x+i)}{x^2+1}\d x
\]
and, by taking real parts, show that
\[
\int_{-∞}^∞\frac{\log\left(x^2+1\right)}{x^2+1}\d x=2π\log2
\]
Solution.
$\operatorname{Log}(i(z+i))-i\fracπ2$ is a branch of $\log(z+i)$ which is holomorphic in the cut-plane region.
Consider the contour $γ=[-R,R]∗γ_R$ where $γ_R(t)=Re^{it},t∈[0,π],R>1$.
$f(z)=\frac{\log(z+i)}{z^2+1}$ has a simple pole $z=i$ inside $γ$. $$\res{z=i}f(z)=\left.\frac{\log(z+i)}{z+i}\right|_{z=i}=\frac{π-2i\log2}4$$By Residue Theorem, the integral of $f$ along $γ$ is $2πi⋅\frac{π-2i\log2}4=π\log2+\frac{iπ^2}2$.
Since $\lim_{z→∞}\abs{\log(z+i)\over z^{1/2}}=0$, there exists $C$ such that $\abs{\log(z+i)}≤C\abs{z}^{1/2}$, then $\abs{f(z)}≤C\abs{z}^{-3/2}$, by Estimation Lemma $∫_{γ_R}f(z)\d z≤πR\abs{f(z)}→0$ as $R→∞$. So \[ \int_{-∞}^∞\frac{\log(x+i)}{x^2+1}\d x=π\log2+\frac{iπ^2}2 \] Taking real parts, since $\Re\left(\log\left(x+i\right)\right)=\log\abs{x+i}=\frac12\log\left(x^2+1\right)$, we get\[ \int_{-∞}^∞\frac{\log\left(x^2+1\right)}{x^2+1}\d x=2π\log2 \] - Show that
\[
\int_0^∞\frac{\sin px\sin qx}{x^2}\d x=\frac{π\min(p,q)}2
\]
where $p, q>0$.
Solution 1.
Using $2\sin px\sin qx=\cos\abs{p-q}x-\cos(p+q)x$, we are left to prove $∀m>0,∫_0^∞\frac{1-\cos(mx)}{x^2}\d x=\frac{πm}4$. By Q2 we know $∀m>0,∫_0^∞\frac{\sin(mx)}x\d x=\fracπ4$.
Integrating by parts, \begin{align*} \int_0^∞\frac{1-\cos(mx)}{x^2}\d x&=\left.-\frac{1-\cos(mx)}x\right|_0^∞+m\int_0^∞\frac{\sin(mx)}x\d x\\ &=m\int_0^∞\frac{\sin(mx)}x\d x\\ &=\frac{πm}4 \end{align*} Solution 2.
Let $a=\max(p,q),b=\min(p,q)$. \begin{align*}2\int_0^∞\frac{\sin(ax)\sin(bx)}{x^2}\d x &=\int_0^∞\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\d x \\ &= \int_0^∞\int_{a-b}^{a+b}\frac{\sin(xy)}x\d y\d x\\ &=\int_{a-b}^{a+b}\int_0^∞\frac{\sin(xy)}{x}\d x \d y\\ &=\int_{a-b}^{a+b}\fracπ2\d y\\&=πb\end{align*} - Let $a∈ℂ$ with $-1<\Re a< 1$. By considering a rectangular contour with corners at $R, R+iπ,-R+iπ,-R$, show that
\[
\int_{-∞}^∞\frac{e^{ax}}{\cosh x}\d x=π\sec\left(\frac{πa}2\right)
\]
and hence evaluate, for real $n$,
\[
\int_{-∞}^∞\frac{\cos n x}{\cosh x}\d x
\]
Solution.
$\cosh z=0⇔z=\frac{i(2k+1)π}2,k∈ℤ$. So $f(z)=\frac{e^{az}}{\cosh z}$ has a simple pole $\frac{iπ}2$ inside the contour. \[ \res{z=\frac{iπ}2}f(z)=\frac{e^{iπa/2}}{\sinh\left(\frac{iπ}2\right)}=\frac{e^{iπa/2}}i \] By Residue Theorem, the integral of $f$ along the contour is $2πi⋅\frac{e^{iπa/2}}i=2πe^{iπa/2}$……(1)
Integrating on each segment, \[ \int_{-R}^Rf(x)\d x+i\int_0^πf(R+ix)\d x-\int_{-R}^R{\color{green}f(x+iπ)}\d x-i\int_0^πf(-R+ix)\d x \] Applying ${\color{green}f(x+iπ)}=-e^{iπa}f(x)$, \[ (1+e^{iπa})\int_{-R}^Rf(x)\d x+i\int_0^πf(R+ix)\d x-i\int_0^πf(-R+ix)\d x\tag2 \] We will prove the 2nd and 3rd term both tend to 0 as $R→∞$. Let $\Re a=α$. Then $\abs{e^{aR}}=e^{αR}$. So $$\abs{f(R+ix)}=\frac{e^{αR}}{\abs{\cosh(R+ix)}}≤\frac{e^{αR}}{\sinh R}$$ Since $0< α< 1$, $$\lim_{R→∞}\frac{e^{αR}}{\sinh R}=0$$ By Estimation Lemma $∫_0^πf(R+ix)\d x→0$ as $R→∞$. Similarly $∫_0^πf(-R+ix)\d x→0$ as $R→∞$. Comparing (1)(2) we have$$\int_{-∞}^∞f(x)\d x=\frac{2πe^{iπa/2}}{1+e^{iπa}}=π\sec\left(πa\over2\right)$$For real $n$, since $\Re(in)=0$, we can apply the above formula,\[ \int_{-∞}^∞\frac{\cos n x}{\cosh x}\d x=\int_{-∞}^∞\frac{e^{inx}+e^{-inx}}{2\cosh x}\d x=π\frac{\sec\left(iπn\over2\right)+\sec\left(-iπn\over2\right)}2=π\operatorname{sech}\left(πn\over2\right) \] - Let $f:ℂ→ℂ$ be continuous. For $ϵ≥0$ consider the paths
\[
γ_{ϵ,R}:[0,R]→ℂ,γ_{ϵ,R}(t)=t+iϵ
\]
If $γ$ is the path $γ_{0,R}$ show that
\[
\int_{γ_{ϵ,R}}f(z)\d z→\int_γf(z)\d z
\]
as $ϵ→0$. (We will be using this exercise when we consider keyhole contours).
Proof.
By Heine–Cantor theorem, $f$ is uniformly continuous on $[0,ϵ]×[0,R]$, so $∀a>0,∃δ>0$, for any $z,w∈[0,ϵ]×[0,R]$ such that $\abs{z-w}< δ$, $\abs{f(z)-f(w)}< a$. For $ϵ∈[0,δ]$, \begin{align*} \abs{\int_{γ_{ϵ,R}}f(z)\d z-\int_γf(z)\d z}&=\abs{\int_0^Rf(x+iϵ)-f(x)\d x}\\ &≤\int_0^R\abs{f(x+iϵ)-f(x)}\d x\\ &≤\int_0^Ra\d x\\ &=aR \end{align*} tends to 0 as $a→0$. - Let $f$ be an entire injective function. Show that $f(z)=az+b$ for some $a,b∈ℂ$.
Proof.
$f$ cannot be a polynomial of degree greater than $1$, by the fundamental theorem of algebra. If $f$ is entire but not a polynomial, then $f(1/z)$ has an essential singularity at $z=0$. By Casorati-Weierstrass, $f(\{z:\abs z>n\})$ is dense in ℂ for each positive integer $n$. By the open mapping theorem, the set is open. By sheet2Q12, $D=⋂_n f(\{z:\abs z>n\})$ is not empty, and every element of $D$ has infinitely many preimages under $f$, contradict. So $f$ is a linear function. - Suppose that $f$ is a holomorphic function defined on an open set $U$ of the complex plane containing $\bar{B}(0,1)$. Let $S^1=\{z∈ℂ:\abs z=1\}=∂B(0,1)$. Show that if $f(S^1)$ is an ellipse and $f$ restricted on $S^1$ is injective, then $f$ is injective on $\bar{B}(0,1)$.
Proof.
1) For $w_0∈ℂ∖f(S^1)$, let $N$ be the number of $z∈B(0,1)$ such that $f(z)=w_0$, then $N$ is the number of zeros of $g(z)=f(z)-w_0$ in $B(0,1)$.
By the argument principle, since $g$ is holomorphic, $N=I(g(S^1),0)=I(f(S^1),w_0)$ which is either 1 (if $w_0$ is inside the ellipse) or 0 (if outside).
2) If $f$ maps a point $z_1$ inside $S^1$ to a point on $f(S^1)$. Since the outside of $S^1$ is open, $∃r:B(z_1,r)⊂\text{inside of }S^1$, by the open mapping theorem, $f(B(z_1,r))$ is open, so $∃w_1∈f(B(z_1,r))∩\text{Exterior of }f(S^1)$, so $w_1$ has a preimage in $B(z_1,r)$, which is inside $S^1$, contract 1). - Suppose that $f: U→ℂ$ is a holomorphic function on an open set $U⊆ℂ$, and suppose $f'(a) ≠0$ for some $a∈U$. Show that
a) there is an $r>0$ such that $f$ is injective on $B(a,r)$
b) its inverse $g$ is given on the image of such a disk by \[ g(w)=\frac1{2πi}\int_γ\frac{zf'(z)}{f(z)-w}\d z \] where $γ(t)=a+re^{it},t∈[0,2π]$.
Proof.
a) If $a$ is a limit point of $\{z∈U∣f(z)-f(a)=0\}$, by Identity Theorem, $f=f(a)$ on $U⇒f'(a)=0$, contradict.
Therefore $∃R > 0:f(z)-f(a)≠0\;∀z∈\bar B(a,R)∖\{a\}$. Then $\abs{f(z)-f(a)}>0$ on $γ(a,R)$.
Since $γ(a,R)$ is compact, $m≔\min\{\abs{f(z)-f(a)}:z∈γ(a,R)\}>0$.
By definition of $m$, $∀z∈γ(a,R),w∈B(f(a),m):\abs{f(z)-f(a)}>\abs{f(a)-w}$.
Since $f(z)-f(a)$ has 1 zero in $B(a,R)$, Rouché's theorem implies that $f(a)-w+f(z)-f(a)=f(z)-w$ has 1 zero in $B(a,R)$.
Since $f$ is continuous at $a$, $∃0< r< R$ such that $f(B(a,r))⊂B(f(a),m)$, so $f|_{B(a,r)}$ is injective.
b) Let $g$ be the inverse of $f$ on $f(B(a,r))$. For $w∈f(B(a,r))$, let $h(z)=f(z)-w$, by Argument Principle, since $h$ has a simple zero at $z=g(w)$,\[g(w)=\frac1{2πi}\int_γz⋅\frac{f'(z)}{f(z)-w}\d z=\frac1{2πi}\int_γz⋅\frac{h'(z)}{h(z)}\d z=g(w) \]