Could anyone help me decide what contour to use to evaluate this integral?

$$\int_0^\infty \frac{\sqrt{t}}{1+t^2} dt$$

So we have simple poles at $i$,$-i$. Why does using a quarter of a circle in the upper right quadrant not work -is it a problem having $i$ on the contour of integration?

My second idea was to let $$\sqrt{t}=e^{\frac{1}{2}\operatorname{log}t}$$

And defining the branch of the logarithm to be the negative imaginary axis. Then using an upper semi-circular contour. With a hole at 0, which will tend to the point 0 as the radius of the semi circle tends to infinity. Then we'll divide by two to get the integral from $0$ to $\infty$. Does this work, as this does not seem to be an even function we're integrating.

I apolygise for the poor explanation of my contours, it's quite difficult without pen and paper!

EDIT: Ok, so using an upper semi circle, with a semicircular hole about $0$. let us call the contour $\gamma$, we have:

$$\int_\gamma \frac{e^{\frac{1}{2}\operatorname{log}t}}{1+t^2}dt =2\pi i\Big(\frac{e^{\frac{1}{2}\operatorname{log}t}}{t+i}\Big)\Big|_{t=i}=\frac{e^{\pi/4}}{2i}$$

But by letting on the positive real axis $z=x+iy$ implies $log(z)=log(x)+0$ and on the negative axis $log(z)=log(x)+\pi$ Therefore if we tend the large arc's radius to infinity and the small arc about the origin's radius to zero we find the integral along the arc tends to zero. So

$$\int_\gamma \frac{e^{\frac{1}{2}\operatorname{log}t}}{1+t^2}dt=\int_0^\infty \frac{\sqrt{x}}{1+x^2} dx+\int_0^\infty \frac{\sqrt{x}e^{\pi/2}}{1+x^2} dx=(1+e^{\pi/2})\int_0^\infty \frac{\sqrt{x}}{1+x^2} dx$$

$$\Rightarrow \int_0^\infty \frac{\sqrt{x}}{1+x^2} =\frac{e^{\pi/4}}{2i(1+e^{\pi/2})}$$

This is wrong but can anyone point out what mistake I have made here?


Call $\displaystyle K=\int_0^\infty \frac{\sqrt{t}}{1+t^2} \mathrm dt$ and $\displaystyle J_\gamma=\int_\gamma\frac{\sqrt{t}}{1+t^2} \mathrm dt$. Basically, you made the same mistake twice.

(1) Evaluating $J_\gamma$ thanks to the residue theorem yields $$ J_\gamma=2\pi\mathrm i\cdot\left.\left(\frac{\mathrm e^{(\log t)/2}}{t+\mathrm i}\right)\right|_{t=\mathrm i}=\pi\mathrm e^{(\log \mathrm i)/2}=\pi\mathrm e^{\mathrm i\pi/4}=\pi(1+\mathrm i)/\sqrt2. $$ Here the mistake is to believe that (the principal value of) $\log\mathrm i$ is $\pi/2$ instead of $\mathrm i\pi/2$: consider that $\mathrm e^{\mathrm i\pi/2}=\cos(\pi/2)+\mathrm i\sin(\pi/2)=\mathrm i$ and that $\mathrm e^{\pi/2}$ is... well, a real number close to $4.81$.

(2) Evaluating $J_\gamma$ through an integral on the real line yields $$ \int_{-\infty}^\infty \frac{\sqrt{t}}{1+t^2} \mathrm dt=K+\int_{-\infty}^0 \frac{\sqrt{t}}{1+t^2} \mathrm dt=K+\int_0^{+\infty} \frac{\mathrm i\sqrt{t}}{1+t^2} \mathrm dt=(1+\mathrm i)\,K. $$ Here, the mistake is to believe that, for every positive real number $t$, $\sqrt{-t}$ is $\mathrm e^{\pi/2}\sqrt{t}$ instead of $\mathrm e^{\mathrm i\pi/2}\sqrt{t}=\mathrm i\sqrt{t}$.

Correcting steps (1) and (2), your reasoning yields the relation $$ (1+\mathrm i)K=\lim\limits_{R\to+\infty}J_{\gamma}=\pi(1+\mathrm i)/\sqrt2, $$ hence the (exact) value $K=\pi/\sqrt2$.

Note: A "purely real" road is available to compute $K$. To do that, first use in succession the changes of variables $u=\sqrt{t}$, $v=\pm u$, and $w=1/v$ to get $$ K=\int_0^\infty\frac{2u^2}{1+u^4}\mathrm du=\int_{-\infty}^\infty\frac{v^2}{1+v^4}\mathrm dv=\int_{-\infty}^\infty\frac{\mathrm dw}{1+w^4}. $$ Thus, $$ K=\frac12\int_{-\infty}^\infty\frac{x^2+1}{x^4+1}\mathrm dx. $$ The factorisation $x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$ implies that $$ \frac{x^2+1}{x^4+1}=\frac1{(\sqrt2x+1)^2+1}+\frac1{(\sqrt2x-1)^2+1}, $$ hence $K=\frac12(L_1+L_{-1})$ where, for every real number $a$, $$ L_a=\int_{-\infty}^\infty\frac{\mathrm dx}{(\sqrt2x+a)^2+1}. $$ Using the change of variable $u=\sqrt2x+a$, one sees that, for every real number $a$, $$ L_a=\frac1{\sqrt2}\int_{-\infty}^\infty\frac{\mathrm du}{u^2+1}=\frac{\pi}{\sqrt2}, $$ where the last equality is direct if one recognizes the density of a standard Cauchy distribution. This proves finally that $K=\pi/\sqrt2$.

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    $\begingroup$ This is very enlightening i'll tell you! That was such a stupid mistake, thank you for pointing it out, I am definitely on the right track now. $\endgroup$ – Freeman Jan 9 '12 at 23:37
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    $\begingroup$ yes, a helpful explanation, thx @Did, also for the link to the Cauchy distribution, which is initially counterintuitive, so very instructive. nice question, too, OP. $\endgroup$ – David Holden Aug 12 '14 at 20:25

Your contour would work if you took into account the behavior of $\sqrt{z}$: your larger integral is not twice the original, but $1+i$ times it.

More generally, your integral and others $\int_0^\infty {t^s\,dt \over 1+t^2}$ (and with other denominators...) can be integrated by the Hankel or keyhole contour: come along the positive reals from $+\infty$ to near $0$, go counter-clockwise around $0$ on a small circle, then back out to $+\infty$ along the positive reals. The trick is that with $s$ non-integral this gives $(1-e^{-2\pi i s})$ times the original integral. Then use residues at the zeros of the denominator.

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    $\begingroup$ Dear Paul: Just in case you didn't see this question. (This a question about one of your texts. I posted an answer. If it is incorrect and/or silly, please feel free state it clearly. Thanks in advance.) $\endgroup$ – Pierre-Yves Gaillard Jan 14 '12 at 5:11

What contour, do you ask? Well, the best contour is no contour at all! :-) All integrals of the form

$\displaystyle\int_0^\infty\frac{t^{^{n-1}}}{a^{^m}+t^{^m}}dt$ can be shown to equal $a^{^{n-m}}\cdot\dfrac\pi m\cdot\csc\bigg(n\cdot\dfrac\pi m\bigg)$, where $\csc x=\dfrac1{\sin x}$, in the

following manner: First, let $t=au$, then, after factoring $a^{^{n-m}}$ outside the integral, let $x=\dfrac1{1+u^{^m}}$

and recognize the expression of the beta function in the resulting integral. Afterwards, apply Euler's

reflection formula for the $\Gamma$ function, in order to arrive at the afore mentioned result. Identifying

the various parameters, we have $I=\dfrac\pi{\sqrt2}$ . $\big($I'm posting this in the answer section because it's too long for a comment $)$.


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