# Using the Residue Theorem for a contour integral along the Riemann sphere

Given the integral $\int_{-\infty}^{\infty}\frac{x}{x^2+1}dx,$ we can clearly see this is the integral of an odd function with limits which are symmetric about the origin, and thus its integral is zero.

However, if I treat this as a curve along the real axis on the Riemann sphere (since the function is zero at infinity), then I can consider the interior of the curve to be the upper-half of the complex plane where it has a single singularity of order $1$ at $i$. Thus, applying the residue theorem, I obtain:

$$\int_{-\infty}^{\infty}\frac{x}{x^2+1}dx = 2\pi i\lim_{x\rightarrow i}(x-i)\frac{x}{(x-i)(x+i)} = 2\pi i\frac{i}{2i} = \pi i\neq 0.$$

Clearly I'm doing something wrong, can someone explain to me my misconception(s)?

Thanks

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Hmm.. But $\pi i$ for sure??? That's rather weird.. –  Berci Feb 15 '13 at 1:16
Well not for sure, it's possible I calculated the residue incorrectly. –  heat death Feb 15 '13 at 1:22
If anything I feel I'm interpreting the concept of a closed curve along the Riemann Sphere incorrectly. –  heat death Feb 15 '13 at 1:26
Well, that real improper integral is not convergent anyway. I think that's all. However I'm not expert in Riemann Sphere integrals.. –  Berci Feb 15 '13 at 1:27
You're probably right Berci. –  heat death Feb 15 '13 at 1:49

It is not true that the integral from -∞ to ∞ of odd function is always 0.

The upper half plane method is not applicable in this question because the integral along the circle with radius appoaching infinity is not 0.

So basicly both of your approaches are incorrect. This integral is the expected value of t(lamda)-distribution with lamda = 1, which is not defined.

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By your first statement do you just mean that we generally don't consider double sided improper integrals defined if a single side does not converge? –  heat death Feb 15 '13 at 1:50
your integral is a typical example. The upper limit of your integral(∞) and the lower limit(-∞) is not the same, therefore, we do not know which ∞ is larger or simply not defined. In the bounded case, if the two bounds are the same real number, the integral is 0 because the function spans symmetrically. –  Shu Xiao Li Feb 15 '13 at 1:55
I'm not sure I quite understand what you mean by we don't know which is larger, can we not just define it to be: $\lim_{t\rightarrow\infty}\int_{-t}^{t}\frac{x}{x^2+1}dx$, which equals zero. –  heat death Feb 15 '13 at 1:57
@cat, that is true for finite intervals, but the integral in question is improper. The integrals of the positive and negative parts of the integrand both diverge, so the integrand is not Lebesgue integrable. By saying it is $0$ you are implicitly taking the principle value of the integral, which is one of many possible values you may assign it. –  Antonio Vargas Feb 15 '13 at 1:57
@cat you could likewise define the integral to be $\lim_{t \to \infty} \int_{-t}^{2t} x/(x^2+1)dx = \log 2$. Why prefer one over the other? The fact is that the integral is divergent in the standard sense so we may choose a value to assign it if we want to. –  Antonio Vargas Feb 15 '13 at 2:08

The integral is zero. The closed contour integral should equal $i\pi$ but the integral over the semi circle is not zero. Set $x=r\text{e}^{i\theta}$, and $0\leq \theta \leq \pi$, and $r\rightarrow \infty$. Then $dx = ir\text{e}^{i\theta} d\theta$, integrate over theta from 0 to $\pi$, and remember $\frac{1}{1+\Delta} = 1 -\Delta + \Delta^2 - \Delta^3...$ and you will find the integral over the arc equals $i\pi$. Since the total integral should also equal $i\pi$ the intgral along the real x axis is zero as for ALL ODD FUNCTIONS. Infinity may be defined differently but you should just do the integral for symmetric cases, instead of giving up because the question wasnt framed unmistakably.

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I think the cheating is that $$\int_0^\infty \frac x{x^2+1}=+\infty$$ does not converge, and the symmetry manipulation just yields $\infty-\infty$ then, which is undefined.

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Epong! Entirely correct. –  ncmathsadist Feb 15 '13 at 2:05

This is my understanding.

Although $\frac{x}{x^2+1}$ is well behaved at $x = \infty$, $dx$ isn't. This is because $x$ itself cannot be used as a complex coordinate on neighborhood of $x = \infty$ of the Riemann sphere.

Let us look at the same integral using a proper complex coordinate $z = \frac{1}{x}$ there. We have: $$\frac{xdx}{x^2+1} \to -\frac{dz}{z(z^2+1)}$$ A $\frac{1}{z}$ singularity becomes apparent in the new coordinate. To properly use the Cauchy Integral theorem, we cannot complete the contour directly at $z = 0$. Instead, we have to circle around it with a small half circle (clockwisely). The contour integral pick up a factor $\pi i = -(-\pi i)$ from this small half circle around $z = 0$ from the $-\frac{1}{z}$ pole.

Translating this back into $x$ world, we cannot complete the contour directly at $x = \infty$. Instead, we have to circle around it with a large half circle (counter-closewisely w.r.t the origin). Since the integrand $\frac{x}{x^2+1}$ doesn't go to $0$ fast enough, the integral pick up a factor $\pi i$ from the large half circle.

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I don't have any background in Riemann surfaces, could you explain your change of coordinates? I can't make sense of it. –  heat death Feb 15 '13 at 4:07
@cat Look at the case of real sphere $S^2$. It can be viewed as two copies of unit closed balls glued together along their boundaries. Same thing happens to the Riemann sphere. It can be viewed as two copies of complex unit disks glued at their unit circles. Let's $x$ and $z$ be the complex coordinates for the two copies of the disk. To glue these two disks, one need to specify how a point with $x$ coordinate on one disk is attached to point with $z$ cooridate on another disk. $z = \frac{1}{x}$ is the canonical choice to glue these two disks together to form the Riemann sphere. –  achille hui Feb 15 '13 at 4:50
But are you just substituting $\frac{1}{z}$ in for $x$? Because I get $\frac{z}{z^2+1}$. –  heat death Feb 15 '13 at 5:48
@cat, $dx = d(\frac{1}{z}) = -\frac{dz}{z^2}$ –  achille hui Feb 15 '13 at 5:50
oh ok I get it. –  heat death Feb 15 '13 at 6:32