Calculate the integral $\int\limits_{0}^{\infty }\frac{x^{s}}{\sqrt{x^2+1}}dx$

calculusreal-analysis

There is a beautiful identity, which I proved (I will not give the proof, because it is not about the identity)
$$\int\limits_{0}^{\infty }x^{s-1}f(x)dx=\Gamma (s)\int\limits_{0}^{\infty }\frac{1}{x^s}\mathcal{L}^{-1}\left \{ f(t) \right \}(x)dx$$
This identity shows a direct connection between the Mellin transform and the inverse Laplace transform

With this identity you can solve integrals quite easily. Yes, if you noticed, the Bessel function plays an important role here!

Calculate the integral
$$\int\limits_{0}^{\infty }\frac{x^{s}}{\sqrt{x^2+1}}dx$$

Here is my attempt to calculate this integral using this identit:

$$\int\limits_{0}^{\infty }\frac{x^{s}}{\sqrt{x^2+1}}dx=\int\limits_{0}^{\infty }\frac{J_0(x)}{x^{-2s+1}}dx=\frac{\pi^{1/2}}{2\sin \pi s}\frac{\Gamma (s+1/2)}{\Gamma (s+1)}=\frac{\pi^{1/2}}{4s\sin \pi s}\Gamma \left ( s-\frac{1}{2} \right )$$

Did I apply this identity correctly? Did I calculate this integral correctly?

Best Answer

I believe the correct derivation is as follows:

$$\mathcal{M}_x\left[\frac{1}{\sqrt{x^2+1}}\right](s+1)=\int\limits_0^{\infty} \frac{x^s}{\sqrt{x^2+1}}\,dx$$ $$=\Gamma(s+1) \int\limits_0^{\infty} x^{-s-1} \left(\mathcal{L}_t^{-1}\left[\frac{1}{\sqrt{t^2+1}}\right](x)\right)\,dx$$ $$=\Gamma(s+1) \int\limits_0^{\infty} x^{-s-1}\, J_0(x)\,dx=\frac{\Gamma\left(-\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right)}{2 \sqrt{\pi}},\quad -1<\Re(s)<0$$

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[Math] Integral $\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx $

It turns out that the Fourier transform of $J_0^3$ can still be expressed in terms of complete elliptic integrals, but it's considerably more complicated than the formula for ${\cal FT}(J_0^2)$: for starters, it involves the periods of a curve $E$ defined over ${\bf C}$ but (except for a few special values of $\omega$) not over ${\bf R}$.

Assume $|\omega| < 3$, else $I(\omega) = 0$. Then the relevant curve is $$ E : Y^2 = X^3 - \bigl(\frac{3}{4} f^2 + \frac{27}{2} f - \frac{81}{4}\bigr) X^2 + 9 f^3 X $$ where $$ f = \frac12 \bigl( e + 1 + \sqrt{e^2-34e+1} \bigr) $$ and $$ e = \bigl( |\omega| + \sqrt{\omega^2-1} \, \bigr)^2. $$ Let $\lambda_1, \lambda_2$ be generators of the period lattice of $E$ with respect to the differential $dx/y$ (note that these are twice the periods that gp reports, because gp integrates $dx/2y$ for reasons coming from the arithmetic of elliptic curves). Then: if $|\omega| \leq 1$ then $$ I(\omega) = \left|\,f\,\right|^{5/2}\, \left|\,f-1\right| \frac{\Delta}{(2\pi)^2}, $$ where $\Delta = \bigl|{\rm Im} (\lambda_1 \overline{\lambda_2}) \bigr|$ is the area of the period lattice of $E$. If $1 \leq |\omega| \leq 3$ then $$ I(\omega) = \left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2} \frac{\Delta'}{(2\pi)^2}, $$ where $\Delta' = \bigl| {\rm Re}(\lambda_1 \overline{\lambda_2}) \bigr|$ for an appropriate choice of generators $\lambda_1,\lambda_2$ (these "appropriate" generators satisfy $|\lambda_1|^2 = \frac32 |\lambda_2|^2$, which determines them uniquely up to $\pm$ except for finitely many choices of $\omega$).

The proof, alas, is too long to reproduce here, but here's the basic idea. The Fourier transform of $J_0$ is $(1-\omega^2)^{-1/2}$ for $|\omega|<1$ and zero else. Hence the Fourier transforms of $J_0^2$ and $J_0^3$ are the convolution square and cube of $(1-\omega^2)^{-1/2}$. For $J_0^2$, this convolution square is supported on $|\omega| \leq 2$, and in this range equals $$ \int_{t=|\omega|-1}^1 \left( (1-t^2) (1-(|\omega|-t)^2) \right)^{-1/2} \, dt, $$ which is a period of an elliptic curve [namely the curve $u^2 = (1-t^2) (1-(|\omega|-t)^2)$], a.k.a. a complete eliptic integral. For $J_0^3$, we likewise get a two-dimensional integral, over a hexagon for $|\omega|<1$ and a triangle for $1 \leq |\omega| < 3$, that is a period of the K3 surface $$ u^2 = (1-s^2) (1-t^2) (1-(|\omega|-s-t)^2). $$ (The phase change at $|\omega|=1$ was already noted here in a now-deleted partial answer.) In general, periods of K3 surfaces are hard to compute, but this one turns out to have enough structure that we can convert the period into a period of the surface $E \times \overline E$ where $\overline E$ is the complex conjugate.

Now to be honest I have only the formulas for the "correspondence" between our K3 surface and $E \times \overline E$, which was hard enough to do, but didn't keep track of the elementary multiplying factor that I claim to be $\left|\,f\,\right|^{5/2}\, \left|\,f-1\right|$ or $\left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2}$. I obtained these factors by comparing numerical values for the few choices of $\omega$ for which I was able to compute $I(\omega)$ to high precision (basically rational numbers with an even numerator or denominator); for example $I(2/5)$ can be computed in gp in under a minute as

intnum(x=0,5*Pi,2*cos(2*x/5) * sumalt(n=0,besselj(0,x+5*n*Pi)^3))

There were enough such $c$, and the formulas are sufficiently simple, that they're virtually certain to be correct.

Here's gp code to get $e$, $f$, $E$, and generators $\lambda_1,\lambda_2$ of the period lattice:

e = (omega+sqrt(omega^2-1))^2
f = (sqrt(e^2-34*e+1)+(e+1)) / 2
E = ellinit( [0, -3/4*f^2-27/2*f+81/4, 0, 9*f^3, 0] )
L = 2*ellperiods(E)
lambda1 = L[1]
lambda2 = L[2]

NB the last line requires use of gp version 2.6.x; earlier versions did not directly implement periods of curves over $\bf C$.

For $\omega=0$ we have $e=1$, $f=3$, and $E$ is the curve $Y^2 = X^3 - 27 X^2 + 243 X = (X-9)^3 + 3^6$, so the periods can be expressed in terms of beta functions and we recover the case $\nu=0$ of Question 404222, How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$? .

Calculate the integral $\int\limits_{0}^{1}\frac{x^{1-p}(1-x)^p}{(1+x)^2}dx$

Substitute $x=\frac1{1+y}$ \begin{align} I=&\int_{0}^{1}\frac{x^{1-p}(1-x)^p}{(1+x)^2}dx =\int_0^\infty\frac{y^p}{(1+y)(2+y)^2}dy\\ =& \int_0^\infty \frac{y^p}{1+y}dy-\int_0^\infty\frac{y^p}{2+y}dy- \int_0^\infty\frac{y^p}{(2+y)^2}dy \end{align} Utilize the known integral $J(a)=\int_0^\infty \frac {y^p}{a+y}dy= -\pi a^p \csc\pi p$ to evaluate \begin{align} & \int_0^\infty \frac{y^p}{1+y}dy=J(1)= -\pi \csc\pi p\\ & \int_0^\infty \frac{y^p}{2+y}dy=J(2)= -\pi \ 2^p\csc\pi p\\ & \int_0^\infty \frac{y^p}{(2+y)^2}dy=-\frac{d J(a)}{da}\bigg|_{a=2} = \pi \ 2^{p-1}p\csc\pi p\\ \end{align} Plug above results into $I$ to obtain $$I=\pi\ [2^{p-1}(2-p)-1]\csc\pi p $$

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[Math] Integral $\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx $

Calculate the integral $\int\limits_{0}^{1}\frac{x^{1-p}(1-x)^p}{(1+x)^2}dx$

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