Problem 2. Calculate an improper integral $I=\int_{0}^{\infty} \frac{\sqrt{x}dx}{(x^2+2x+1)}$.
(a) Calculate $ J_{R,\epsilon}=\int_{\Gamma_{R,\epsilon}} f(z)dz, \ f(z):= \frac{\sqrt{z}}{(z^2+2z+1)}$
$\Gamma_{R,\epsilon}$ is the contour on the figure.
(b)Prove that $\int_{\gamma_R} f(z)dz \rightarrow 0$ and $\int_{\gamma_ \epsilon} f(z)dz \rightarrow 0$ as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$ where $\gamma_R$ and $\gamma_\epsilon$ are arcs.
(c) Express limit of $J_{R,\epsilon}$ as $R \rightarrow +\infty$, $\epsilon \rightarrow 0^+$ using $I$.
Answer:
(a) Since $f(z) = \frac{\sqrt{z}}{(z^2+2z+1)} = \frac{\sqrt{z}}{(z+1)^2}$, $z=-1$ is the only singularity inside $\Gamma_{R,\epsilon}$ as $R>1$.
The residue is $Res(f(z),-1)=\frac{(\sqrt{z})'}{1!}|_{z=-1} = \frac{\frac{1}{2\sqrt{z}}}{1!}|_{z=-1} = -\frac{i}{2}$
Thus, by residue theorem $J_{R,\epsilon}= 2\pi i Res(f(z),-1)=2\pi i (-\frac{i}{2})=\pi$
(b) \begin{align*}
\int_{\gamma_R} f(z)dz &\leq \left| \int_{\gamma_R} f(z)dz \right|\\
&\leq 2\pi R\cdot max \left|\frac{\sqrt{z}}{(z^2+2z+1)} \right|\\
&\leq 2\pi R\cdot max \left| \frac{R^{\frac{1}{2}}e^{i\frac{1}{2}t}}{(Re^{it}+1)^2}\right| \ \ \ , \ \text{$z=Re^{it}, t\in [0,2\pi]$}\\
&\leq 2\pi R \cdot \frac{R^{\frac{1}{2}}}{(R+1)^2} \rightarrow 0 \ \ \ \ as \ R \rightarrow \infty\\
\\
\int_{\gamma_ \epsilon} f(z)dz &\leq \left|\int_{\gamma_ \epsilon} f(z)dz\right|\\
&\leq 2\pi \epsilon \cdot max \left| \frac{\sqrt{z}}{(z^2+2z+1)} \right|\\
&\leq 2\pi \epsilon \cdot max \left| \frac{\epsilon^{\frac{1}{2}}e^{i\frac{1}{2}t}}{(\epsilon e^{it}+1)^2}\right| \ \ \ , \ \text{$z=\epsilon e^{it}, t\in [2\pi,0]$}\\
&\leq 2\pi \epsilon \cdot max (\frac{\epsilon^{\frac{1}{2}}}{(|1|-|\epsilon e^{it}|)^2})\\
&\leq 2\pi \epsilon \cdot \frac{\epsilon^{\frac{1}{2}}}{(1-\epsilon)^2} \rightarrow 0 \ \ \ \ as \ \epsilon \rightarrow 0\\
\end{align*}
(c)\begin{align*}
J_{R,\epsilon} &= \int_{\gamma_ R} f(z)dz +\int_{\gamma_ \epsilon} f(z)dz +\int_{\epsilon}^{\infty} f(z)dz+\int_{\infty}^{\epsilon}f(z)dz\\
\pi &= 0+0+\int_{\epsilon}^{\infty} f(z)dz+\int_{\infty}^{\epsilon}f(z)dz\\
\end{align*}
As $R \rightarrow +\infty$, $\epsilon \rightarrow 0^+$,
\begin{align*}
\int_{\epsilon}^{\infty} f(z)dz &= \int_{0}^{\infty}\frac{\sqrt{x}dx}{(x^2+2x+1)}=I\\
\int_{\infty}^{\epsilon}f(z)dz &= \int_{\infty}^{0}\frac{\sqrt{z}}{(z^2+2z+1)}dz \ \ \text{, $z=xe^{i2\pi},dz=e^{i2\pi} dx$}\\
&=\int_{\infty}^{0}\frac{\sqrt{x}e^{i\pi}}{(xe^{i2\pi}+1)^2} e^{i2\pi}dx\\
&=\int_{0}^{\infty}\frac{\sqrt{x}}{(x+1)^2}dx\\
&=I
\end{align*}
Then, the limit of $J_{R,\epsilon}$ as $R \rightarrow +\infty$, $\epsilon \rightarrow 0^+$ is $2I$. Thus, $2I=\pi \Rightarrow I=\frac{\pi}{2}$
