\text{Major II}
\text{Sommerfeld-Watson transformation}
\text{Branch-cuts and integration}
\text{Linear ODEs}
\text{Beta and Gamma functions}
\text{ Evaluate the following integral using complex integration techniques.}\\
\displaystyle I(p)=\int_0^{+\infty} \frac{x^p}{1+(x+1)^2} d x,
\text{Problem 1:}
\displaystyle \oint f(z) d z=2 \pi i \sum \operatorname{Res}(f(z))
(z+1)^2-i^2=0
z_1=-1+i, z_2=-1-i
\displaystyle \operatorname{Res}\left(f(z), z_1\right)=\frac{z_1^p}{z_1-z_2}=\frac{e^{3 i \frac{\pi}{4} p}}{2 i} 2^{\frac{p}{2}}
\displaystyle \operatorname{Res}\left(f(z), z_2\right)=\frac{z_2^p}{z_2-z_1}=\frac{e^{\frac{5\pi i}{4} p}}{-2 i} 2^{\frac{p}{2}}
\displaystyle \oint f(z) d z=\pi\left(e^{3 i \frac{\pi}{4} p}-e^{5 i \frac{\pi}{4} p}\right) {2}^{p/2}
\displaystyle \oint f(z) d z=\int_{C_1} f(z) d z+\int_{C_2} f(z) d z+\int_{C_r} f(z) d z+\int_{C_R} f(z) d z
\text{Optional (+0.25 pts): Find the limit when }p\rightarrow 0
\displaystyle \left(1-e^{2 i \pi p}\right) I=\pi\left(e^{3 \frac{ i \pi p}{4}}-e^{\frac{5 i\pi p}{4}}\right) 2^{p / 2}
\displaystyle I=\frac{\pi}{e^{i \pi p}} \frac{e^{i \pi p}\left(e^{-2 i \frac{\pi p}{8}}-e^{+2 i \frac{\pi p}{8}}\right)}{-2 i \sin (\pi p)} 2^{\frac{p}{2}}
\displaystyle I=2^{p / 2} \pi \quad \frac{\sin \left(\frac{\pi p}{4}\right)}{\sin (\pi p)}
\displaystyle \int_{c_2} f(z) d z=\int_{\infty}^0 \frac{x e^{i 2 \pi p}}{(x+1)^2+1} d x=-e^{i 2 \pi p} I
\displaystyle \int_{c_1} f(z) d x=\int f(x) d x=I
\text{2) The limit}
\text{since }\sin x\approx x
\displaystyle I \approx \pi \quad \frac{ \left(\frac{\pi p}{4}\right)}{ (\pi p)}\rightarrow \pi/4
(\text{The integrals on }C_r \text{ and }C_R \text{ vanish at he two,limits })
\text{Problem 2:}
\text{Let us consider the following integral:}
\displaystyle I(p,q)=\int_0^\infty \frac{x^p}{|1-x|^q}dx
\text{1) Separate the integral to two intervals }[0,1]\text{ and }[1,+\infty[
\displaystyle \text{2) Make the change of variable }x=\frac{1}{u} \text{ for the second integral.}
\text{3) Express now } I(p,q) \text{ using Beta functions}
\displaystyle \text{4) Calculate the result for }p=1/4, q=1/2,\text{as function of }\Gamma{(1/4)}\text{ alone }
\text{5) Calculate the limit for } p,q \rightarrow \infty \text{ with the constraint }p-q=\frac{1}{2} \text{ with } p, q \text{ non integrers}
\text{Solution:}
\displaystyle \int_0^{\infty} \frac{x^p}{|1-x|^q} d q=\int_0^1 \frac{x^p}{|1-x|^q} d x+\int_1^{+\infty} \frac{x^p}{|1-x|^q} d x .
\displaystyle u=\frac{1}{x} \Rightarrow d x=-\frac{d u}{u^2}
\displaystyle \int_1^{\infty} \frac{x^p}{|1-x|^q} d x=\int_0^1 \frac{u^{-p}}{|1-\frac{1}{u}|^q} \frac{d u}{u^2}=\int_0^1 \frac{u^{q-p-2}}{|1-u|^q} d u
\displaystyle =B(q-p-1,1-q)
\displaystyle I(p, q)= B(q-p-1,1-q)+B(p+1,1-q)
\displaystyle I(1/4,1/2)= B(-3/4,1/2)+B(5/4,1/2)
\displaystyle I(1/4,1/2)= \frac{\Gamma(-3/4)\Gamma(1/2)}{\Gamma(-1/4)}+\frac{\Gamma(5/4)\Gamma(1/2)}{\Gamma(7/4)}
\displaystyle I(1/4,1/2)= \frac{\Gamma(1/4)^2}{3\sqrt{2\pi}} +\frac{\Gamma(1/4)^2}{3\sqrt{2\pi}}
\Gamma(3/4)\Gamma(1/4)=\pi/\sin(\pi/4)=\pi\sqrt{2}
\Gamma(5/4)=(1/4) \Gamma(1/4)
\Gamma(1/4)=(-3/4) \Gamma(-3/4)
\Gamma(3/4)=(-1/4) \Gamma(-1/4)
\Gamma(z+1)=z \Gamma(z) ;
\displaystyle \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi z}}
\Gamma(7/4)=(3/4) \Gamma(3/4) =\frac{3\pi\sqrt{2}}{(4\Gamma(1/4))}
\displaystyle I(1/4,1/2)= 2 \frac{\Gamma(1/4)^2}{3\sqrt{2\pi}}
\text{Evaluate the following sum:}
\displaystyle S(a)=\sum_{n=-\infty}^{\infty} \frac{1}{(n-a)^2+1}
\text{Problem 3:}
\text{Optional: What is the limit when }a\rightarrow 0
\text{Solution:}
\displaystyle F(z)=\frac{\pi}{(z-a)^2+1} \frac{\cos \pi z}{\sin \pi z}
\text{The poles of }f(z) \text{ are: } z_1=a+i, z_2=a-i
\displaystyle \sum_{z_i} \text{Res}(F(z),z_i)=\frac{\pi}{\left(z_1-z_2\right)} \frac{\cos \pi z_1}{\sin \pi z_1}+\frac{\pi}{\left(z_2-z_1\right)} \frac{\cos \pi z_2}{\sin \pi z_2}
\displaystyle =\frac{\pi}{\left(z_1-z_2\right)}\Bigg( \frac{\cos \pi z_1}{\sin \pi z_1}- \frac{\cos \pi z_2}{\sin \pi z_2}\Big)
\displaystyle =-\pi i \frac{\cos \pi z_1 \sin \pi z_2- \cos \pi z_2 \sin \pi z_1}{2\sin \pi z_1 \sin \pi z_2}
\displaystyle =-\pi i \frac{\sin \pi (z_2-z_1)}{2\sin \pi z_1 \sin \pi z_2}
\displaystyle =\pi i \frac{\sin 2\pi i }{2\sin \pi z_1 \sin \pi z_2}
\displaystyle \sum_{n=-\infty}^{n=+\infty} f(n)= -\pi i \frac{i\sinh 2\pi }{\cosh 2\pi - \cos 2\pi a}\rightarrow \pi \frac{\sinh 2\pi }{\cosh 2\pi - 1}=\pi \coth(\pi)
\displaystyle =\pi i \frac{\sin 2\pi i }{(\cos \pi (z_1-z_2)- \cos \pi (z_1+z_2))}
\displaystyle =\pi i \frac{\sin 2\pi i }{(\cos 2\pi i - \cos 2\pi a)}
\text{As we can see in this figure, the sum oscillates as function of a.}
\text{Problem 4:}
\text{Solve the following isobaric ODE}
\displaystyle \frac{d y}{d x}=\frac{-\left(y^2+\frac{2}{x}\right)}{2 y x}
\text{Solution:}
\displaystyle 2 y x d y=-\left(y^2+\frac{2}{x}\right) dx
\text{we find that } 2m+1=0\Rightarrow m=-1/2
y=vx^{-1/2}\Rightarrow y^2=v^2 /x \Rightarrow 2 y dy=2vdv/x-v^2 dx/x^2
2vdv-v^2 dx/x =-(v^2/x+2/x) dx\Rightarrow v dv=-dx/x
\text{The equation (1) becomes:}
(1)
\Rightarrow v^2=-2\ln x -2c
\Rightarrow y^2 x=-2\ln cx
\displaystyle \Rightarrow y =\Big(-2\frac{\ln (c/x)}{x}\Big)^{1/2}
\text{Problem 5:}
\text{Let us solve the following \textbf{non-linear} ODE:}
\displaystyle y'+x y=\frac{x^2+x}{y}
\text{1) Make the change of variable }u=1+y^2
\text{2) Solve the ODE with the variable }u
3) \text{Find }y.
\text{Solution}
\displaystyle y'+x y=\frac{x^2+x}{y}
u=1+y^2\Rightarrow u'=2y y'
u'+2x(u-1)=2(x^2+x)
\Rightarrow u'+2 x u =2x^2+4x
u(x)=\lambda e^{-x^2}
\text{the homogeneous solution is }
\text{variation of the constant}
\lambda' e^{-x^2}-2x \lambda e^{-x^2}+2x \lambda e^{-x^2}=2x^2+4x
\lambda'= e^{x^2}(2x^2+4x)
\displaystyle \lambda= \int^x e^{s^2}(2s^2+4s) ds =F(x)
u(x) =F(x)e^{-x^2}
y(x) =(F(x)e^{-x^2}-1)^\frac{1}{2}
\text{Let us solve the following differential equation: }
y^{\prime \prime}+y^{\prime}-2 y=e^x+x
\text{Problem 6:}
\text{1) Find the solution for the homogeneous equation.}
\text{2) Find the Wronskian. }
\text{3) Find the particular solution. }
\text{4) Express the general solution.}
\text{The characteristic polynomial is:}
m^2+m-2=0
\text{Solution:}
\text{the roots are: } m_1=1,m_2=-2
\text{the solutions for the homogeneous equation are:}
e^x \text{ and } e^{-2x}
\displaystyle
W(x)=
\begin{vmatrix}
e^x & e^{-2x} \\
e^x & -2e^{-2x}
\end{vmatrix}
=-2e^{-x}-e^{-x}=-3e^{-x}
\displaystyle y_p(x)=-\frac{1}{3}\Big(e^{-2x}\int^x e^{2s} F(s) ds-e^x\int^x e^{-s} F(s) ds \Big)
\displaystyle y_p(x)=-\frac{1}{3}\Big(e^{-2x}\int^x (e^{3s}+s e^{2s}) ds-e^x\int^x (1-se^{-s}) ds \Big)
\displaystyle y_p(x)=-\frac{1}{3}\Big( \Big((e^{x}/3+(e^{-2x}+(2x-1))/4) \Big) -(xe^x -(e^x +(x+1))) \Big)
y(x)=A e^x+Be^{-2x}+y_p(x)
1+i=\sqrt{2}e^{i\pi/4}
-1+i=\sqrt{2}e^{i3\pi/4}
-1-i=\sqrt{2}e^{i5\pi/4}
1-i=\sqrt{2}e^{i7\pi/4}
\displaystyle e^{i\theta_1}+e^{i\theta_2}=2e^{i\frac{\theta_1+\theta_2}{2}} \cos{\Bigg(\frac{\theta_1-\theta_2}{2}\Bigg)}
\displaystyle e^{i\theta_1}-e^{i\theta_2}=2ie^{i\frac{\theta_1+\theta_2}{2}} \sin{\Bigg(\frac{\theta_1-\theta_2}{2}\Bigg)}
\displaystyle \cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2}
\displaystyle \sin{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2i}
i=e^{i\pi/2}
-1=e^{i\pi}
z=|z|e^{i\text{arg}(z)}
2\Re(z)=z+\overline z
2i\Im(z)=z-\overline z
\displaystyle \Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin \pi z}
\displaystyle \Gamma\Big(\frac{1}{2}\Big)=\sqrt{\pi}
\Gamma(z) \Gamma\left(z+\frac{1}{2}\right)=2^{1-2 z} \sqrt{\pi} \Gamma(2 z)
\displaystyle \Gamma(z) \equiv \int^{\infty}_{0} e^{-t} t^{z-1} d t,\hspace{0.15 cm} \text{Re}(z)>0
\displaystyle \Gamma(z+1) \sim \sqrt{2 \pi z}\left(\frac{z}{e}\right)^z, z\rightarrow \infty
\displaystyle \mathcal{B}(a, b)=\frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}
\displaystyle y_p(x)=y_2(x) \int^x \frac{y_1(s) F(s) d s}{W\left\{y_1(s), y_2(s)\right\}}-y_1(x) \int^x \frac{y_2(s) F(s) d s}{W\left\{y_1(s), y_2(s)\right\}}
\displaystyle \mathcal{B}(x, y)=\int_0^1 t^{x-1}(1-t)^{y-1} d t
\displaystyle \mathcal{B}(x, y)=2 \int_0^{\pi / 2}(\sin \theta)^{2 x-1}(\cos \theta)^{2 y-1} d \theta
\text { for } x, y>0
\displaystyle \int x e^{a x} dx=\frac{1+e^{a x}(a x-1)}{a^2}
\displaystyle F(z)=\pi f(z) \frac{\cos (\pi z)}{\sin (\pi z)}
\displaystyle \sum_{n=-\infty}^{n=+\infty} f(n)=-\sum_i \operatorname{Res}\left(F(z), z=z_i\right)
\displaystyle \cos(i x)=\cosh(x)
\displaystyle \sin(i x)=i \sinh(x)
\sin(x)\approx x
\cos(x)\approx 1-x^2/2