TT2 Problem 4

[Victor Ivrii]

$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Calculate an improper integral
$$
I=\int_0^\infty \frac{dx}{\sqrt{x}(x^2+1)}.
$$

Hint:

(a)  Calculate
$$
J_{R,\varepsilon} = \int_{\Gamma_{R,\varepsilon}} f(z)\,dz, \qquad f(z)=\frac{1}{\sqrt{z}(z^2+1)}
$$
where we have chosen the branch of $\sqrt{z}$ such that it is analytic on the upper half-plane $\{z\colon \Im z>0\}$ and is real-valued for $z=x>0$. $\Gamma_{R,\varepsilon}$ is the contour on the figure below:

(b) Prove that $\int_{\gamma_R}  \frac{dz}{\sqrt{z}(z^2+1)}\to 0$ as $R\to \infty$ and $\int_{\gamma_\varepsilon}  \frac{dz}{\sqrt{z}(z^2+1)}\to 0$ as $\varepsilon\to 0^+$ where $\gamma_R$ and $\gamma_\varepsilon$ are large and small semi-circles on the picture. This will give you a value of
$$
\int_{-\infty}^0 f(z)\,dz + \int_0^{\infty} f(z)\,dz.
$$

(c) Express both integrals using $I$.

[Junya Zhang]

Following the hint, consider $f(z)=\frac{1}{\sqrt{z} (z^2 + 1)}$ and contour $\gamma$ as shown in given picture.
Then $$\int_{\gamma}{f(z)dz}
=\int^{-\epsilon}_{-R}{f(x)dx}
+ \int^{R}_{\epsilon}{f(x)dx}
+  \int_{\gamma_{\epsilon}}{f(z)dz}
+  \int_{\gamma_{R}}{f(z)dz} $$

Step 1
Fix $0 < \epsilon < 1$ and $R>1$.
Then $f$ is analytic everywhere in $\gamma$ except at $z = i$, where it has a simple pole.
Then by the residue theorem, $$\int_{\gamma}{f(z)dz} = 2\pi i \left[\frac{1}{\sqrt{z}(z+i)}\right]_{z=i} = \frac{2\pi i}{2i\sqrt{i}} = \frac{\pi }{\sqrt{i}} $$ $$= \frac{\pi }{\sqrt{e^{i\pi/2}}} = \frac{\pi }{e^{i\pi/4}} =\frac{\pi }{\cos(\pi/4) + i \sin(\pi/4)} = \frac{\pi }{\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}}
= \frac{\pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)}
= \frac{\pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)}{\frac{1}{2}+ \frac{1}{2}}
= \pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)$$

Overcomplicated, but correct. V.I
Note that this holds for all $0 < \epsilon < 1$ and $R>1$.

Step 2
$$\int^{-\epsilon}_{-R}{f(x)dx}  = \int^{-\epsilon}_{-R}{\frac{1}{\sqrt{x} (x^2 + 1)}dx} = \int^{\epsilon}_{R}{\frac{1}{\sqrt{-x} (x^2 + 1)} (-dx)} = \int^{R}_{\epsilon}{\frac{1}{\sqrt{-x} (x^2 + 1)}dx} = \frac{1}{i} \int^{R}_{\epsilon}{\frac{1}{\sqrt{x} (x^2 + 1)}dx} $$

Step 3
$$\int^{R}_{\epsilon}{f(x)dx}  = \int^{R}_{\epsilon}{\frac{1}{\sqrt{x} (x^2 + 1)}dx} $$

Step 4
Consider $\left\lvert{\int_{-\gamma_{\epsilon}}{f(z)dz}}\right\lvert$. With the parametrization $z = \epsilon e^{i\theta}$ for $\theta\in[0,\pi]$.
$$\left\lvert{\int_{-\gamma_{\epsilon}}{f(z)dz}}\right\lvert
= \left\lvert{\int_{0}^{\pi}{\frac{i\epsilon e^{i\theta}}{\sqrt{\epsilon e^{i\theta}} ((\epsilon e^{i\theta})^2 + 1)}d\theta}}\right\lvert
\leq {\int_{0}^{\pi}{
\frac{\left\lvert i\epsilon e^{i\theta}\right\lvert}
{\left\lvert\sqrt{\epsilon e^{i\theta}}\right\lvert  \left\lvert (\epsilon e^{i\theta})^2 + 1\right\lvert}
d\theta}}
= \int_{0}^{\pi} \frac{\epsilon}{\sqrt{\epsilon} \left\lvert (\epsilon e^{i\theta})^2 + 1\right\lvert}d\theta
\leq \int_{0}^{\pi} \frac{\epsilon}{\sqrt{\epsilon} (1 - \epsilon^2)}d\theta
= \frac{\pi\epsilon}{\sqrt{\epsilon} (1 - \epsilon^2)}
$$
$$\lim_{\epsilon \to 0^+}{\frac{\pi\epsilon}{\sqrt{\epsilon} (1 - \epsilon^2)}} = \lim_{\epsilon \to 0^+}{\frac{\pi\sqrt{\epsilon}}{ (1 - \epsilon^2)}}  = \frac{0}{1} = 0$$
This implies that $$\lim_{\epsilon \to 0^+}{\left\lvert{\int_{-\gamma_{\epsilon}}{f(z)dz}}\right\lvert} = 0 $$
Which then gives $$\lim_{\epsilon \to 0^+}{{\int_{-\gamma_{\epsilon}}{f(z)dz}}} = 0 $$ and thus $$\lim_{\epsilon \to 0^+}{{\int_{\gamma_{\epsilon}}{f(z)dz}}} = -0 = 0 $$

Step 5
Consider $\left\lvert{\int_{\gamma_{R}}{f(z)dz}}\right\lvert$. With the parametrization $z = R e^{i\theta}$ for $\theta\in[0,\pi]$.
$$\left\lvert{\int_{\gamma_{R}}{f(z)dz}}\right\lvert
= \left\lvert{\int_{0}^{\pi}{\frac{iR e^{i\theta}}{\sqrt{R e^{i\theta}} ((R e^{i\theta})^2 + 1)}d\theta}}\right\lvert
\leq {\int_{0}^{\pi}{
\frac{\left\lvert iR e^{i\theta}\right\lvert}
{\left\lvert\sqrt{R e^{i\theta}}\right\lvert  \left\lvert (R e^{i\theta})^2 + 1\right\lvert}
d\theta}}
= \int_{0}^{\pi} \frac{R}{\sqrt{R} \left\lvert (R e^{i\theta})^2 + 1\right\lvert}d\theta
\leq \int_{0}^{\pi} \frac{R}{\sqrt{R} (R^2-1)}d\theta
= \frac{\pi R}{\sqrt{R} (R^2-1)}
$$
$$\lim_{R \to \infty}{\frac{\pi R}{\sqrt{R} (R^2 - 1)}} = \lim_{R \to \infty}{\frac{\pi}{\sqrt{R} (R - \frac{1}{R})}}  = \frac{\pi}{\infty \cdot \infty} = 0$$
This implies that $$\lim_{R \to \infty}{\left\lvert{\int_{\gamma_{R}}{f(z)dz}}\right\lvert} = 0 $$
Which then gives $$\lim_{R \to \infty}{{\int_{\gamma_{R}}{f(z)dz}}} =  0 $$

Altogether, as $\epsilon \to 0^+$ and $R \to \infty$, the equation $\int_{\gamma}{f(z)dz}
=\int^{-\epsilon}_{-R}{f(x)dx}
+ \int^{R}_{\epsilon}{f(x)dx}
+  \int_{\gamma_{\epsilon}}{f(z)dz}
+  \int_{\gamma_{R}}{f(z)dz} $ becomes
$$ \pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right) = \frac{1}{i} I  + I + 0 + 0 $$
Equating the real part of the equation gives $$\frac{\pi}{\sqrt{2}} = I$$
Checking: equating the imaginary part of the equation gives $$-\frac{i\pi}{\sqrt{2}} = \frac{1}{i}I$$
which gives $$I = \frac{\pi}{\sqrt{2}}$$ as well.

[Andong Liu]

(a) $x^{2} + 1 = 0$
so $x = i$ $(x = -i$ not in range of graph )
By Residue theorem
$\int f\left(z\right)dz = 2\pi i \sum Res(f, i)= (\dfrac{1}{\sqrt{z}(z + i)}, i) \times 2\pi i = \dfrac{1}{\sqrt{i}(2i)}\times 2\pi i = \dfrac{\pi}{\sqrt{i}} $
(b)  $|\int_{R} \dfrac{dz}{\sqrt{z}(z^{2} + i)}| \leq \pi R  \dfrac{1}{\sqrt{R}(R^{2} - i)}\rightarrow 0$ When $R\rightarrow \infty$
$|\int_{\varepsilon} \dfrac{dz}{\sqrt{z}(z^{2} + i)}| \leq \pi \varepsilon  \dfrac{1}{\sqrt{\varepsilon}(i - \varepsilon^{2})} \rightarrow 0$ When $\varepsilon\rightarrow 0$
so, we have $\int ^{0}_{-\infty} f\left(z\right)dz + \int ^{\infty}_{0} f\left(z\right)dz = \dfrac{\pi}{\sqrt{i}} $
(c) $\int ^{0}_{-\infty} f\left(z\right)dz + \int ^{\infty}_{0} f\left(z\right)dz = 2I = \dfrac{\pi}{\sqrt{i}} $
$I = \dfrac{\pi}{2 \sqrt{i}} $

[Junya Zhang]

@liuandon
We are computing a real integral (integral over real values), the result should not have any $i$ in it.

[Victor Ivrii]

Andong
Incorrectly expressed $\int_{-\infty}^0$ via $I$ (not taken in account $\arg{\sqrt{x}}$ as $x<0$