Use residue theorem to calculate the integral $\int_{0}^{\infty} \frac{x^{1/2}}{1+x^2} dx$

contour-integrationresidue-calculus

I'm working through Boas Chapter 14, and I came across an interesting issue in problem 33.

I get a value for the integral $I = \int_{0}^{\infty} \frac{x^{1/2}}{1+x^2} dx = \frac{i\pi\sqrt{2}}{2}$. While the answer in the book (and elsewhere online including this website) has no $i$ component, and everything works just fine.

This question has been asked before: Evaluate $\int_{0}^{\infty} \frac{x^{1/2}}{1 + x^2}\,\mathrm dx$ using Residue Theorem. but the answer on that question from 2013 has an error in the calculation of the Residue at $-i$.

Specifically, the answer claims

$$
\text{Res}_{-i} \frac{\sqrt{z}}{z^2+1} = \frac{\sqrt{-i}}{-2i} = \frac{-1+i}{-2i\sqrt{2}} = \frac{1-i}{2i\sqrt{2}}
$$
However
$$
\sqrt{-i} = (e^{-i\pi/2})^{1/2} = e^{-i\pi/4} = \cos{\frac{\pi}{4}} – i\sin{\frac{\pi}{4}} = \frac{1}{\sqrt{2}} -\frac{i}{\sqrt{2}}
$$
Thus
$$
\frac{\sqrt{-i}}{-2i} = \frac{1-i}{-2i\sqrt{2}} = \frac{-1+i}{2i\sqrt{2}}
$$

With this correction, the sum of the residues is $\frac{\sqrt{2}}{2}$, and therefore the value of the contour integral $i\pi\sqrt{2}$.

If you carry out the rest of the contour integration, you pick up the $1/2$ term, and end up with $I = \frac{i\pi\sqrt{2}}{2}$.

Am I missing something here? I saw elsewhere online people mistakenly calculating the Residue at $-i$, is there another way to lose this imaginary component of the answer?

EDIT:

My friend pointed out my error – incase anyone else makes the same mistake as me.

The polar angle must be measured in the positive direction. Meaning

$$
-i = e^{3\pi/2} \qquad \text{and not}\qquad -i=e^{-\pi/2}
$$

Then, when you apply the square root, you end up with

$$
-i = e^{3\pi/4} = \frac{-1 + i}{\sqrt{2}}
$$

Which eliminates the imaginary component in the sum of the residues, and therefore in the final answer.

Best Answer

You seem to be using the wrong branch for the square root. We have $-i=e^{3\pi i/2}$; note that because of how this integral is being calculated (using a "pacman contour" with pacman's mouth along the non-negative real axis), in the residue calculus we should use the branch where the argument lies between $0$ and $2\pi$. So, \begin{align} \text{Res}\left(\frac{\sqrt{z}}{z^2+1};-i\right) = \frac{\sqrt{-i}}{-2i}=\frac{e^{3\pi i/4}}{2e^{3\pi i /2}} = \frac{e^{-3\pi i/4}}{2}=\frac{-e^{i\pi/4}}{2}=\frac{-1-i}{2\sqrt{2}}, \end{align} which is the same thing as $\frac{1-i}{2i\sqrt{2}}$ as in the other answer.

Related Solutions

[Math] Integral with contour integration

You could also use the following contour. contour

$$\int_{-\infty}^0 \frac{2x^2-1}{x^4+1}\operatorname dx =\int_{0}^{+\infty} \frac{2x^2-1}{x^4+1}\operatorname dx $$

Has an analytic continuation as $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz $$ with 4 poles, but just on pole inside of the contour.

$$\operatorname*{res}_{z=e^{i\frac{\pi}{4}}} f(z) = \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8} $$

I think you made a calculation error here (?)

Using $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz = \color{blue}{\int_{\Gamma_1}\frac{2z^2-1}{z^4+1}\operatorname dz} + \int_{\Gamma_2}\frac{2z^2-1}{z^4+1}\operatorname dz + {\color{red}{\int_{\Gamma_3}\frac{2z^2-1}{z^4+1}\operatorname dz}}$$

Now $\color{blue}{\int_{\Gamma_1} \to 0}$ as $R\to +\infty$ which can be proven using the triangle inequality.
Use $\Gamma_2 \leftrightarrow z(x) = x$ and $x:0\to R$
Use $\color{red}{\Gamma_3 \leftrightarrow z(y) = iy}$ and $y: R \to 0$

Which finally results in (as $R \to +\infty$) $$2\pi i \left(\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\right) = \color{blue}{0}+\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx + \color{red}{ i \int_{+\infty}^0\frac{-2y^2-1}{y^4+1}\operatorname dy}$$

Here you can read the real parts which results in:

$$\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx = \frac{\pi\sqrt{2}}{4}$$

An integral with residue theorem

Let us consider the following contour.

Fix $R>0$ a radius which is a ("big") multiple of $2\pi$, $R=R(n)=2\pi n$. Consider the closed contour $C=C(R)$ build from the two pieces

the segment (interval) $I=I(R)$ from $-R$ to $R$,
the halfcircle $H=H(R)$ from $R$ to $-R$ parametrized by $t\to Re^{it}$, $t\in[0,\pi]$.

A picture would be:

complex analysis contour integral controlling <span class= $1+e^x$ in the denominator">

and the unit is $\pi$. Then we have for the given meromorphic function $f(z)= (e^z+1)^{-1}(z^2+\pi^2)^{-1}$ $$ \begin{aligned} \int_{\Bbb R}f(x)\; dx & = \lim_{n\to\infty} \int_{-R(n)}^{+R(n)} f(z)\; dz = \lim_{n\to\infty} \int_{I(R(n))} f(z)\; dz \ , \\ 0&= \lim_{n\to\infty} \int_{C(R(n))} f(z)\; dz \\ \int_{I(R)} f(z)\; dz + \int_{H(R)} f(z)\; dz &= 2\pi i\sum_{a\text{ pole inside }C(R)} \operatorname{Residue}_{z=a}f(z) \ . \end{aligned} $$

The zero limit, when integrating on $C(R(n))$, $n\to\infty$ can be motivated as follows. The contour integral is a closed set, not passing through the poles. There is a minimal distance $\pi$ to the poles, which helps us to bound from below $|1+e^z|$, so from above $1/|1+e^z|$. Then the term $1/(x^2+\pi^2)$ is in $O(R^{-2})$, the contour length $(\pi R)$ in $O(R)$, so we land in $O(1)\cdot O(R^{-2})\cdot O(R^1)=O(R^{-1})$.

It remains to consider the sums of the residues of the poles in the upper half plane. We get: $$ \begin{aligned} \operatorname{Residue}_{z=i\pi}f(z) &= \operatorname{Residue}_{h=0\\z=i\pi+h} \frac 1{1+e^{i\pi}e^h}\cdot \frac 1{h+i\pi+i\pi}\cdot \frac 1{h+i\pi-i\pi} \\ &= \operatorname{($h^0$-Coefficient)}_{h=0} \frac 1{1-e^h}\cdot \frac 1{h+2\pi i} \\ &= \operatorname{($h^0$-Coefficient)}_{h=0} \frac 1{-h-\frac 12h^2+O(h^3)}\cdot \frac 1{2\pi i}\cdot \frac 1{1-(-h/(2\pi i))} \\ &= \operatorname{($h^0$-Coefficient)}_{h=0} -\frac 1h \left(1-\frac 12h+O(h^2)\right) \frac 1{2\pi i}\cdot \left(1-\frac h{2\pi i}+O(h^2)\right) \\ &= \operatorname{($h^0$-Coefficient)}_{h=0} -\frac 1{2\pi i}\cdot \frac 1h \left(1-\frac 12h-\frac h{2\pi i}+O(h^2)\right) \\ &= \frac 1{2\pi i}\cdot \left(\frac 12+\frac 1{2\pi i}\right) \end{aligned} $$ I really wanted to do this with bare hands, and use computer only for the check:

sage: var('z');
sage: f(z) = 1 / (1+exp(z)) / (z^2+pi^2)
sage: f(z).residue( z==i*pi )
-1/4*I/pi - 1/4/pi^2
sage: bool( f(z).residue( z==i*pi ) == 1/(2*pi*i) * (1/2 + 1/(2*pi*i)) )
True

It turns out, that all other residues (at poles in the upper half plane) are real. This time we let the computer give us some values:

sage: f.residue( z==3*pi*i )
z |--> 1/8/pi^2
sage: f.residue( z==5*pi*i )
z |--> 1/24/pi^2
sage: f.residue( z==7*pi*i )
z |--> 1/48/pi^2
sage: f.residue( z==9*pi*i )
z |--> 1/80/pi^2
sage: f.residue( z==11*pi*i )
z |--> 1/120/pi^2

and so on as in the OP, the denominators in the rational numbers appearing above (and touching $1/\pi^{2}$) are squares of odd integers, taken minus one.

Our integral is real, so we can forget all these values, getting thus: $$ \int_{\Bbb R} \frac{dx}{(1+e^x)(x^2+\pi^2)} = 2\pi i \left[ \frac 1{2\pi i} \left( \color{blue}{\frac 12}+\frac 1{2\pi i}\right) + \text{real number(s)}\right] \ , $$ and only the blue term survives to give us the "real answer".

Note: I was spending a lot of time trying to check / validate somehow this value numerically. This is the best i could get in pari/gp:

? \p 1000
   realprecision = 1001 significant digits (1000 digits displayed)
? intnum( x=-200, +100, 1 / (1+exp(x)) / (x^2+Pi^2) )
%19 = 0.49500041117264675924354715963413516...

(And we can tacitly control the piece from $[-\infty, -200)$ using $1/(x^2+\pi^2)$.)

Best Answer

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[Math] Integral with contour integration

An integral with residue theorem

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