7
$\begingroup$

I'm trying to show that $\int_1^{\infty}\frac{dx}{x\sqrt{x^2-1}}=\frac{\pi}{2}$ by letting $f(z)=\frac{1}{z\sqrt{z^2-1}}=\frac{1}{z}e^{-\log(z^2-1)^{\frac{1}{2}}}$.

and I need to show that, on the 4 straight lines, $$L_1^+=\{z|z:R+\epsilon i \rightarrow \rho+1+\epsilon i\}, L_1^-=\{z|z:\rho+1-\epsilon i \rightarrow R-\epsilon i\}, $$$$L_2^+=\{z|z:-\rho-1+\epsilon i \rightarrow -R+\epsilon i\}, L_2^-=\{z|z:-R+\epsilon i \rightarrow -\rho-1+\epsilon i\}$$, which are the parts of a simple closed contour which is surrounding the simple pole $z=0$ of $f$ and excluding $\{x\in R | x$ is less than or equal to $-1$ or $x$ is greater than or equal to $1 \}$, $$\lim_{\rho\rightarrow 0,R\rightarrow \infty, \epsilon\rightarrow 0 }\int_{L_{1,2}^{+,-}}f(z)dz=-\frac{\pi}{2}.$$ it is negative since I chose the contour negatively oriented at the first. I was able to show that the other parts go to $0$, but somehow those four line integrals cancel out each other and make me crazy.

specifically, I got the following results:

$L_1^+ \rightarrow -\frac{\pi}{2}$

$L_1^- \rightarrow \frac{\pi}{2}$

$L_2^+ \rightarrow \frac{\pi}{2}$

$L_2^- \rightarrow -\frac{\pi}{2}$

I think the problem is to choose new branch cut when I make $\epsilon$ go to $0$. if $z=\sigma e^{i \phi}$ and $(z^2-1)^\frac{1}{2}=r e^{i\theta}$, then $2\theta=Arg(\sigma^2 e^{2i \phi}-1)$, so $\theta$ behaves similarly with $\phi$, and from here the above results come out and I don't know where I did wrong.

I know the other methods like letting $u=\sqrt{x^2-1}$, but I just want to do it this way to get used to it. any helps or hints? thanks in advance.

$\endgroup$
2
$\begingroup$

Here is the contour of integration and the signs of $\sqrt{z^2-1}$ near the real axis. For large $|z|$, in the upper half-plane, $\sqrt{z^2-1}\approx+z$, whereas in the lower half-plane, $\sqrt{z^2-1}\approx-z$. Note that along the imaginary axis, $\sqrt{z^2-1}$ is positive imaginary, so at $z=0$, we have $\sqrt{z^2-1}=+i$.

This means that $\operatorname*{Res}\limits_{z=0}\left(\frac1{z\sqrt{z^2-1}}\right)=-i$.

enter image description here

The integral along the blue contours vanishes. On the large blue circles, the size of the integrand is $\sim\frac1{|z|^2}$ and on the small blue circles, the size of the integrand is $\sim\frac1{\sqrt{2|z-1|}}$ and $\sim\frac1{\sqrt{2|z+1|}}$

The integral along each of the red and green contours is the integral we want due to the signs of $z$, $\sqrt{z^2-1}$, and the direction of the contour.

Therefore, $$ \begin{align} 4\int_1^\infty\frac{\mathrm{d}z}{z\sqrt{z^2-1}} &=2\pi i\operatorname*{Res}_{z=0}\left(\frac1{z\sqrt{z^2-1}}\right)\\ &=2\pi \end{align} $$ Thus, $$ \int_1^\infty\frac{\mathrm{d}z}{z\sqrt{z^2-1}}=\frac\pi2 $$

$\endgroup$
  • $\begingroup$ Very nice answer. (+1). I somehow missed the fact that the target integral appears twice on the left branch cut. $\endgroup$ – Marko Riedel Oct 9 '16 at 17:44
1
$\begingroup$

The way I would treat this is using

$$f(z) = \frac{1}{z} \exp((-1/2)\mathrm{LogA}(z+1)) \exp((-1/2)\mathrm{LogB}(z-1))$$

with $\mathrm{LogA}$ the branch with argument from $-\pi$ to $\pi$ and $\mathrm{LogB}$ the branch with argument from $0$ to $2\pi$. The first logarithm has branch cut $(-\infty, -1)$ on the negative real axis and the second one from $[1,\infty)$ on the positive real axis.

We use the contour integral

$$ \int_\Gamma f(z) dz$$

where $\Gamma$ consists of six components namely a circle $\Gamma_0$ $1+\epsilon e^{i\theta}$ centered at one with $0\lt\theta\lt 2\pi$ traversed clockwise, a line segment $\Gamma_1$ just above the cut from $\mathrm{LogB}$ connecting to $\Gamma_0$ and continuing up to the point $R$ on the positive real axis, a quarter circle $\Gamma_2$ parameterized by $Re^{i\theta}$ with $0\lt\theta\lt \pi/2$ traversed counterclockwise and connecting $\Gamma_1$ to the point $iR$ on the imaginary axis, a line segment $\Gamma_3$ connecting $iR$ to $-iR$ on the imaginary axis, a quarter circle $\Gamma_4$ parameterized by $Re^{i\theta}$ with $-\pi/2\lt \theta\lt 0$ traversed counterclockwise and finally a line segment $\Gamma_5$ just below the cut of $\mathrm{LogB}$ connecting $\Gamma_4$ to $\Gamma_0.$

Call the target integral $J$ i.e.

$$J = \int_1^\infty \frac{1}{x\sqrt{x^2-1}} dx.$$

Now along $\Gamma_1$ we get

$$f(z) = \frac{1}{x} \exp((-1/2)\log(x+1)) \exp((-1/2)\log(x-1)) = \frac{1}{x\sqrt{x^2-1}}$$

while on $\Gamma_5$ we obtain

$$f(z) = \frac{1}{x} \exp((-1/2)\log(x+1)) \exp((-1/2)\log(x-1)+(-1/2)2\pi i) \\ = \exp(-\pi i))\frac{1}{x\sqrt{x^2-1}} = -\frac{1}{x\sqrt{x^2-1}}.$$

Therefore the contribution from $\Gamma_1$ and $\Gamma_5$ in the limit is $2J.$ We then obtain

$$2J = 2\pi i \times \frac{1}{2} \mathrm{Res}_{z=0} f(z)$$

if we can show that the remaining contributions vanish/cancel. Note however that

$$\mathrm{Res}_{z=0} f(z) = \exp((-1/2)\mathrm{LogA}(1))\exp((-1/2)\mathrm{LogB}(-1)) \\ = \exp((-1/2)\pi i) = -i$$

so that this yields

$$\bbox[5px,border:2px solid #00A000]{ J = \pi i\times \frac{1}{2} \times (-i) = \frac{\pi}{2}.}$$

which is the desired result. (Here we have used the property of $\Gamma_3$ being a line segment passing through the pole at $z=0$ and hence picking up half the residue.)

To conclude we must examine the remaining segments of the contour. To apply the ML bound to $\Gamma_0$ we observe that

$$\left|\frac{1}{1+\epsilon e^{i\theta}} \exp((-1/2)\mathrm{LogA}(2+\epsilon e^{i\theta})) \exp((-1/2)\mathrm{LogB}(\epsilon e^{i\theta}))\right| \\ \le \frac{1}{1-\epsilon} \exp((-1/2)\log(2-\epsilon)) \exp((-1/2)\log(\epsilon))$$

where we have used the fact that the imaginary term produced by the two logarithms being exponentiated has modulus one. ML now yields

$$2\pi\epsilon \times \frac{1}{1-\epsilon} \frac{1}{\sqrt{2-\epsilon}} \frac{1}{\sqrt\epsilon} \rightarrow 0 \quad\text{as}\quad \epsilon\rightarrow 0.$$

For $\Gamma_2$ and $\Gamma_4$ we get (these are alike and we discuss $\Gamma_2$)

$$\left|\frac{1}{1+R e^{i\theta}} \exp((-1/2)\mathrm{LogA}(1 + R e^{i\theta})) \exp((-1/2)\mathrm{LogB}(-1 + R e^{i\theta}))\right| \\ \le \frac{1}{R-1} \exp((-1/2)\log(R-1)) \exp((-1/2)\log(R-1))$$

so that ML yields

$$\frac{1}{4}\times 2\pi R\times \frac{1}{R-1} \frac{1}{\sqrt{R-1}} \frac{1}{\sqrt{R-1}} \\ = \frac{1}{4}\times 2\pi R\times \frac{1}{(R-1)^2} \rightarrow 0 \quad\text{as}\quad R\rightarrow \infty.$$

We now show that the contribution from $\Gamma_3$ is zero where we will traverse clockwise to simplify the notation. We have

$$f(it) = \frac{1}{it} \exp((-1/2)\mathrm{LogA}(it+1)) \exp((-1/2)\mathrm{LogB}(it-1)) \\ = \frac{1}{it} \exp((-1/2)\log\sqrt{t^2+1}+(-1/2)i\mathrm{ArgA}(it+1))\\ \times \exp((-1/2)\log\sqrt{t^2+1}+(-1/2)i\mathrm{ArgB}(it-1)).$$

Observe however that when $\mathrm{ArgA}(it+1)=\theta$ with $-\pi/2\lt\theta\lt\pi/2$ we have $\mathrm{ArgB}(it-1)=\pi - \theta$

so that this simplifies to

$$f(it) = \frac{1}{it} \frac{1}{\sqrt{t^2+1}} \exp((-1/2)i(\theta+\pi-\theta)) = -\frac{1}{t\sqrt{t^2+1}}.$$

Supposing we make a semicircular indentation of radius $\epsilon$ around the singularity at $t=0$ we get for the upper segment parameterized by $z=it$

$$\int_\epsilon^R f(it) i\; dt.$$

We get for the lower segment

$$\int_{-R}^{-\epsilon} f(it) i\; dt = -\int_R^\epsilon f(-iu) i\; du = \int_\epsilon^R f(-iu) i\; du = -\int_\epsilon^R f(iu) i\; du.$$

The two segments cancel for a net contribution of zero. We have examined all six constituents of the closed contour $\Gamma$ and this concludes the argument showing that $J=\frac{\pi}{2}.$

Addendum, Oct 8 2016. In order to justify the segment $\Gamma_3$ passing through the pole at $z_0 = 0$ we observe that a function $f(z)$ with a simple pole at $z_0$ has the form

$$f(z) = F'(z) + \frac{1}{z-z_0} \mathrm{Res}_{z=z_0} f(z)$$

with $F(z)$ analytic in a neighborhood of $z_0.$ Now by making a small semicircular indentation $\Gamma_3'$ that is parameterized by $z=z_0 + \epsilon e^{i\theta}$ with $\theta$ ranging from $\pi/2$ to $-\pi/2$ so as to capture the inner angle we get for the integral of $f(z)$ along the semicircle with $F(z)$ analytic

$$F(z_0-i\epsilon) - F(z_0+i\epsilon) + \mathrm{Res}_{z=z_0} f(z) \times \int_{\Gamma_3'} \frac{1}{z-z_0} dz \\ = F(z_0-i\epsilon) - F(z_0+i\epsilon) + \mathrm{Res}_{z=z_0} f(z) \times \int_{\pi/2}^{-\pi/2} \frac{1}{\epsilon\exp(i\theta)} i\epsilon\exp(i\theta) \; d\theta \\ = F(z_0-i\epsilon) - F(z_0+i\epsilon) + i \mathrm{Res}_{z=z_0} f(z) \times [\theta]_{\pi/2}^{-\pi/2}.$$

This becomes

$$-\pi i \mathrm{Res}_{z=z_0} f(z)$$

as $\epsilon\rightarrow 0.$ We thus obtain

$$2J - \pi i \times (-i) = 2\pi i \times 0$$

or $2J - \pi = 0$ as before. This material is from this MSE link and this German complex variables PDF.

Addendum, Oct 9 2016. The excellent answer by @robjohn states the contributions on the two horizontal segments above and below the left branch cut without proof. It may be interesting to see how these are computed. Starting again from

$$f(z) = \frac{1}{z} \exp((-1/2)\mathrm{LogA}(z+1)) \exp((-1/2)\mathrm{LogB}(z-1))$$

we take a point on the negative real axis $x < -1$ and get from above

$$\frac{1}{x} \exp((-1/2)\log(-x-1)+(-1/2)i\pi) \exp((-1/2)\log(-x+1)+(-1/2)i\pi) \\ = \frac{1}{x\sqrt{(-x)^2-1}} \exp(-i\pi) = - \frac{1}{x\sqrt{x^2-1}}.$$

The contribution is thus

$$- \int_{-\infty}^{-1} \frac{1}{x\sqrt{x^2-1}} dx = - \int_{\infty}^{1} \frac{1}{(-u)\sqrt{u^2-1}} (-1) du \\ = - \int_{\infty}^{1} \frac{1}{u\sqrt{u^2-1}} du = \int_1^{\infty} \frac{1}{u\sqrt{u^2-1}} du = J.$$

On the other hand we get from below

$$\frac{1}{x} \exp((-1/2)\log(-x-1)+(-1/2)(-i\pi)) \exp((-1/2)\log(-x+1)+(-1/2)i\pi) \\ = \frac{1}{x\sqrt{x^2-1}}.$$

The contribution is now

$$\int_{-1}^{-\infty} \frac{1}{x\sqrt{x^2-1}} dx = \int_1^{\infty} \frac{1}{(-u)\sqrt{u^2-1}} (-1) du \\ = \int_1^{\infty} \frac{1}{u\sqrt{u^2-1}} du = J.$$

$\endgroup$
  • $\begingroup$ Does $\Gamma_{3}$ intersect the point $z=0$? I'm having difficulty understanding why that doesn't cause any trouble. $\endgroup$ – JessicaK Oct 8 '16 at 3:27
  • $\begingroup$ This is the Cauchy Principal Value as presented e.g. at this MSE link which says that the contribution from a line through a pole is $\pi i$ times the residue. The contour is indented in the part where I prove that the contribution from $\Gamma_3$ vanishes. $\endgroup$ – Marko Riedel Oct 8 '16 at 4:19
  • $\begingroup$ When you have $f(z)$ on $\Gamma_5$, since $\Gamma_5$ is approaching from the below the real line, shouldn't we take the branch cut of $log(z+1)$ to include $2\pi$, like $\pi < \theta < 3\pi$? the difficulty I'm having here is this; it makes $\Gamma_5$ cancel $\Gamma_1$ out. $\endgroup$ – user159234 Oct 8 '16 at 8:23
  • $\begingroup$ @JessicaK I have added some additional material to address your question. $\endgroup$ – Marko Riedel Oct 8 '16 at 23:59
  • $\begingroup$ @JohnSnow You need to consider the factor resulting from the logarithm on $\Gamma_5,$ it prevents cancellation (factor shown above). $\endgroup$ – Marko Riedel Oct 9 '16 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.