A confusing integral

Question

I'm trying to calculate the integral from one to infinity of the integrand $1/(x*\sqrt{x^2-1})$. On Youtube I've found calculations using integration by substitution, but those calculations don't satisfy me since I prefer to use methods from complex analysis.

My first question is what's going on at the point $x=1$. Is that a pole of the integrand? If so, of what order is it? According to Wolfram Alpha it's a pole of order $1/2$, but that seems strange since I thought that the order of a pole is always an integer. Maybe the concepts of pole and order are not applicable to the given integrand at all since it involves the square root function, which is maybe not meromorphic? What about the concept of residue. Is that applicable to non meromorphic functions? Complex analysis is really a confusing subject.

Furthermore I don't know what contour I should construct. If I draw a vertical line straight up from the point $z=1$ I get a nasty parametrization involving a sum of the form $1 + iy$. So maybe my contour should go partly along the imaginary axis instead, but a problem with that seems to be that the integral doesn't converge between $0$ and $1$, at least not according to Wolfram Alpha.

Can someone clear the confusion and tell me how to proceed?

Maybe the mentioned integral is discussed elsewhere on MSE, but there I can't find answers to my questions about the concepts of pole, order, meromorphic, and residue, so please don't delete what I've written.

Did you mean, from one to infinity? This integral diverges if you take it from zero to infinity — FShrike, May 15 '23 at 14:06
Please format your formulas, and use a more descriptive title. — Anne Bauval, May 15 '23 at 14:14
Hint: if you must use a contour integral, start with a substitution to ensure the integrand has no surd, e.g. $y=\sqrt{x^2-1}$. You can then halve an even integrand & use a semicircular contour. — J.G., May 15 '23 at 14:24
I disagree that this is a duplicate since the op has some specific confusions. The question is not all about : "how do I evaluate the integral?" — FShrike, May 15 '23 at 14:31
Ok then what about this one? Prove that $\int_1^{\infty}\frac{dx}{x\sqrt{x^2-1}}=\frac{\pi}{2}$ using complex analysis — Anne Bauval, May 15 '23 at 14:32
$x=1$ is a branch point for this integrand. (It is not a pole.) — GEdgar, May 15 '23 at 15:54
Maybe this and this are ideas for drawing contours. For your integrand, you can write $\sqrt{x^2-1}=\exp\left(\frac{1}{2}\log(x^2-1)\right)$. There's a branch point, not a pole, at $x=1$ because $1^2-1 = 0$ is a branch point of the logarithm. Since $e^{2\pi i} = e^0$, both $2\pi i$ and $0$ are among many values of $\log(1)$. — Accelerator, May 15 '23 at 18:40

FShrike · Answer 1 · 2023-05-15T14:34:03.640

Did you mean, from one to infinity? This integral diverges if you take it from zero to infinity.

Wolfram just means a generalised statement: if $\lim_{x\to1}(x-1)^{1/2}f(x)$ is nonzero and finite, it is reasonable to say the pole is order $1/2$. However this is not in the sense of complex analysis, where poles only have integer order. The integrand is indeed not meromorphic at $1$.

The concept of residue is "not applicable" directly here since the function fails to be meromorphic in any annulus at $1$ (there is no local Laurent series expansion and hence no meaningful residue). Besides, to use the residue theorem you first need to choose a closed contour.

If you want to evaluate: $$\int_1^\infty\frac{1}{x\sqrt{x^2-1}}\,\mathrm{d}x$$I suggest first taking $x\mapsto\sqrt{x}$ in the integrand so that you only have to deal with: $$\frac{1}{2}\int_1^\infty\frac{1}{x\sqrt{x-1}}\,\mathrm{d}x$$And then using a "keyhole contour" method around the branch cut of square root on $[1,\infty)$.

Sure. @AnneBauval But this was a complex analysis question – FShrike May 15 '23 at 14:31 — FShrike, May 15 '23 at 14:31

score -1 · Answer 2 · answered May 15 '23 at 14:11

-1

The function

f(x) = 1/(xsqrt(x^2-1))*

has a simple pole at x = 1. The order of a pole can only be an integer, so Wolfram Alpha's result of 1/2 must be due to the square root. However, the concept of order still applies to this pole, and it is of order 1.

The residue at x = 1 can be found by using the formula:

Res(f, 1) = lim (x → 1) (x - 1) f(x)

We have:

f(x) = 1/(xsqrt(x^2-1))* *f(x) = 1/[(x-1)sqrt((x-1)^2+2x-1)]

Using L'Hopital's rule, we can evaluate the limit:

Res(f, 1) = lim (x → 1) 1/sqrt((x-1)^2+2x-1) = 1/sqrt(2)

So the residue at x = 1 is 1/sqrt(2).

To evaluate the integral, we can use the residue theorem. Since the integrand has singularities at x = ±1, we can choose a contour that consists of the real axis from 0 to ∞, a small semicircle in the upper half-plane around x = 1, and a large semicircle in the upper half-plane that encloses the entire real axis.

The integral over the large semicircle goes to zero as the radius goes to infinity, since the integrand decays faster than 1/r as r → ∞. The integral over the small semicircle can be evaluated using the residue at x = 1. We have:

∫C f(z) dz = 2πi Res(f, 1)

where C is the contour we have chosen. The integral over the real axis can be evaluated using the Cauchy principal value. We have:

∫-∞^∞ f(x) dx = lim ε → 0+ [∫0^1 f(x) dx + ∫1+ε^∞ f(x) dx - πi Res(f, 1)]

Putting everything together, we get:

∫0^∞ f(x) dx = π/sqrt(2)

So the value of the integral is π/sqrt(2).

answered May 15 '23 at 14:11

Kobrakixx

64

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This is difficult to read. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Martin R May 15 '23 at 14:13
@MartinR oh thanks for suggestion , i wanted that a lot ! – Kobrakixx May 15 '23 at 14:16
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It does not have a simple pole at $x=1$ since $(x-1)f(x)\to0$ here. It does not have a residue at $x=1$ either, as it does not even have a Laurent series in a neighbourhood of $x=1$ – FShrike May 15 '23 at 14:23
@FShrike Thank you for your comment, FShrike. If this limit it zero, doesn't that mean that the functions has a removable singularity at the point? I think a throrem by Riemann suggests that. – A A May 15 '23 at 15:20
@AA But this function is not meromorphic near $1$ so we cannot really talk about such theorems... there is not a removable singularity at $1$ – FShrike May 15 '23 at 15:46
Aside from the mistakes you made for the $x=1$ stuff, you can't draw a contour straight over $x=-1$ because there is another branch point there. Not only that, it doesn't seem very obvious how you would deal with $\int_{-1+\epsilon}^{0}f(z)$ since $f(z)$ diverges as $z \to 0$. – Accelerator May 15 '23 at 18:56

A confusing integral

2 Answers2