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I've been trying to integrate the following

$$\int_{0}^{\infty} \frac{\sqrt{x}}{x^{2}+1} \mbox{d} x$$

on half an annulus in the upper half plane. I keep getting $\frac{\pi}{\sqrt{2}}\ i$, which doesn't give with the numerical approximations I get using WolframAlpha.

How should I attack this problem?

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I guess you want to encircle $x=i$? –  Fabian Jan 20 '13 at 10:50
    
The answer is indeed $\pi/\sqrt2$, hence you should probably just keep track of the factor $i$. –  Did Jan 20 '13 at 10:50
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this is a special case of this problem –  Santosh Linkha Jan 20 '13 at 11:16

4 Answers 4

Though this is not a "complex analysis" solution I can't resist to post it. We have the following identities $$ I=\int\limits_0^\infty\frac{\sqrt{x}}{x^2+1}dx=\{x\to t^2\}=2\int\limits_0^\infty\frac{t^2}{t^4+1}dt=\{t\to t^{-1}\}=2\int\limits_0^\infty\frac{1}{t^4+1}dt $$ Hence $$ I=\int\limits_0^\infty\frac{t^2+1}{t^4+1}dt= \int\limits_0^\infty\frac{d(t-t^{-1})}{(t-t^{-1})^2+2}dt= \int\limits_{-\infty}^\infty\frac{du}{u^2+2}= \frac{1}{\sqrt{2}}\arctan\left(\frac{u}{\sqrt{2}}\right)\Biggl|_{-\infty}^{\infty}=\frac{\pi}{\sqrt{2}} $$

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One way to attack this particular problem is to make the substitution $x=t^2$, as in Ahlfors, Complex Analysis, Third Edition, p. 159; the integral becomes

$$\int_{-\infty}^{\infty} dt \: \frac{t^2}{t^4+1} $$

We may now use a semicircular contour $C$ in the half plane $\Re{z} \ge 0$ rather than the annulus as we have disposed of the branch point at the origin. We write

$$ \oint_C dz \: \frac{z^2}{z^4+1} = i 2 \pi \left ( \mathrm{Res}_{z=e^{i \frac{\pi}{4}}} \frac{z^2}{z^4+1} + \mathrm{Res}_{z=e^{i \frac{3 \pi}{4}}} \frac{z^2}{z^4+1} \right ) $$

$$\begin{align} \mathrm{Res}_{z=e^{i \frac{\pi}{4}}} \frac{z^2}{z^4+1} &= \frac{e^{i \frac{2 \pi}{4}}}{\left (e^{i \frac{\pi}{4}}-e^{i \frac{3 \pi}{4}}\right )\left (e^{i \frac{\pi}{4}}-e^{i \frac{-3 \pi}{4}}\right )\left (e^{i \frac{\pi}{4}}-e^{i \frac{-\pi}{4}}\right )} \\ &= \frac{i}{i (1-i)(2) \left (2 i \sin{\frac{\pi}{4}} \right )} \\ &= -i \frac{1+i}{4 \sqrt{2}} \end{align}$$

Similarly,

$$ \mathrm{Res}_{z=e^{i \frac{3 \pi}{4}}} = -i \frac{1-i}{4 \sqrt{2}} $$

Therefore,

$$ \oint_C dz \: \frac{z^2}{z^4+1} = i 2 \pi (-i) \frac{1}{2 \sqrt{2}} = \frac{\pi}{\sqrt{2}} $$

Now, about the contour $C$ which has a large radius of, say, $R$:

$$ \oint_C dz \: \frac{z^2}{z^4+1} = \int_{C_R} dz \: \frac{z^2}{z^4+1} + \int_{-R}^R dt \: \frac{t^2}{t^4+1} $$

In the limit as $R \rightarrow \infty$,

$$\int_{C_R} dz \: \frac{z^2}{z^4+1} \sim \frac{\pi}{R} $$

and therefore vanishes. We may then conclude that

$$\int_{-\infty}^{\infty} dt \: \frac{t^2}{t^4+1} = \int_0^{\infty} \frac{\sqrt{x}}{x^2+1} = \frac{\pi}{\sqrt{2}} $$

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Pardon me, but if $x=t^2$, shouldn't $\sqrt{x} = t$? Also, shouldn't we get the contour in the upper half plane? –  user58456 Jan 20 '13 at 19:49
    
Yes, but also $dx = 2 t \, dt$, and $\Re{z} \ge 0$ is the upper half plane. –  Ron Gordon Jan 20 '13 at 19:59

(Using the approach suggested by the OP.) Let $\sqrt{z}$ denote the branch of the complex square root with a branch cut along the negative real axis, and look at the indented semi-circular contour $\Gamma$.

enter image description here

$\newcommand{\Res}{\operatorname{Res}}$

Let $f(z) = \dfrac{\sqrt{z}}{1+z^2}$, and integrate along $\Gamma$.

Since $f$ is bounded near the origin, the integral over the small semi-circle tends to $0$ as the radius tends to $0$, and the integral over the large semi-circle $C_R^+$ tends to $0$ by the standard estimation lemma:

$$ \left| \int_{C_R^+} \frac{\sqrt{z}}{1+z^2}\,dz \right| \le \pi R \cdot \frac{2}{R^{3/2}} $$ for $R$ sufficiently large.

The integral over the positive real axis is exactly what we're looking for, and the integral over the negative real axis will be $$ \int_{-\infty}^0 \frac{i\sqrt{-x}}{1+x^2}\,dx = i\int_0^{\infty} \frac{\sqrt{x}}{1+x^2}\,dx. $$

Summing up, using the residue theorem: $$ \int_0^{\infty} \frac{\sqrt{x}}{1+x^2}\,dx + i\int_0^{\infty} \frac{\sqrt{x}}{1+x^2}\,dx = 2\pi i \Res(f; i) = 2\pi i\,\frac{\sqrt{i}}{2i} = \pi\left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt2} \right). $$

Taking the real (or imaginary) part of this final equality gives $$ \int_0^{\infty} \frac{\sqrt{x}}{1+x^2}\,dx = \frac{\pi}{\sqrt{2}}.$$

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This is clearly without using Complex Analysis.

Putting $x=\tan t, x=0\implies t=0$ and $x=\infty, t=\frac\pi2$

$$\int_{0}^{\infty} \frac{\sqrt{x}}{x^2+1}$$

$$=\int_{0}^{\frac\pi2}\sqrt{\tan t}dt=\int_{0}^{\frac\pi2}\sqrt{\tan \left(\frac\pi2+0-t\right)}dt$$ as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

So, $$\int_{0}^{\frac\pi2}\sqrt{\tan t}dt=\int_{0}^{\frac\pi2}\sqrt{\cot t}dt=I\text{ (say),}$$

So, $$2I=\int_{0}^{\frac\pi2}\sqrt{\tan t}dt+\int_{0}^{\frac\pi2}\sqrt{\cot t}dt$$

$$=\int_{0}^{\frac\pi2}\frac{\sin t+\cos t}{\sqrt{\sin t\cos t}}dt$$

Let $\sin t-\cos t=y,$ so $dy=(\cos t+\sin t)dt, y^2=1-2\sin t\cos t, t=0\implies y=-1, t=\frac\pi2 \implies y=1$

So, $$2I=\int_{-1}^1\sqrt2\frac{dy}{\sqrt{1-y^2}}=\sqrt2 (\arcsin y)_{-1}^1=\sqrt2\left\{\frac\pi2-\left(-\frac\pi2\right)\right\}=\sqrt2\pi$$

So, $$I=\frac\pi{\sqrt2}$$

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I think the OP wants to solve the problem using complex integration, since (s)he says "....on half an annulus in the upper half plane". –  Isomorphism Jan 20 '13 at 11:05
    
@Isomorphism, another answer (by Norbert) has also excluded Complex Integration. Btw, did you down-vote? –  lab bhattacharjee Jan 20 '13 at 13:49
    
@labbhattacharjee (+1) –  no identity Jan 20 '13 at 14:14
    
@Norbert, I'm more interested in the mistake with me that has caused the down-voting. –  lab bhattacharjee Jan 20 '13 at 14:21
    
@labbhattacharjee There is no mistake, our answer just unappropriate for the question tagged (complex-analysis) –  no identity Jan 20 '13 at 14:23

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