Integrate $I(x) = \frac{1}{ 2 \pi} \int_{-\infty}^{\infty} \frac{e^{-Na}}{jt(1+jt)} e^{\frac{Na}{1+jt}} e^{jxt}dt $

contour-integrationdefinite integralsimproper-integralsintegrationspecial functions

I want to integrate:

$$I(x) = \frac{1}{ 2 \pi} \int\limits_{-\infty}^{\infty} \frac{e^{-Na}}{jt(1+jt)} e^{\frac{Na}{1+jt}} e^{jxt} \,dt $$

Where $j = \sqrt{-1}$, $a \in \mathbb{R_{++}}$ and $N \in \mathbb{Z_{++}}$

The solution is apparently:

$$ I(x) = e^{-Na} \int_{0}^{x} e^{-y} I_{0}(2 \sqrt{Nay})\,dy $$

where $I_{0}(\cdot)$ is the modified Bessel function of the 1st kind of order 0.

It looks like I should be going for an answer of the form:

$$I(x) = e^{-Na} \big( F(T) – F(0) \big)$$

Where $F(\cdot)$ is the antiderivative of $e^{-x} I_{0}(2 \sqrt{Nax})$ wrt x.

My approach:

We start off by taking out $e^{-Na}$ and writing $e^{\frac{Na}{1+jt}}$ as a power series. This gives us:

$$I(x) = \frac{e^{-Na}}{2 \pi} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{jt(1+jt)} \displaystyle\sum\limits_{k=0}^{\infty} \frac{(Na)^{k}}{k! (1+jt)^{k}} e^{jxt} dt $$

Bringing the sum out and combining gives us:

$$I(x) = \frac{e^{-Na}}{2 \pi} \displaystyle\sum\limits_{k=0}^{\infty}
\frac{(Na)^{k}}{k!}\displaystyle\int\limits_{-\infty}^{\infty} \frac{(Na)^{k}}{jt(1+jt)^{k+1}} e^{jxt} dt $$

This is where I get stuck. The result of the integral is obviously the sum of the residues, but I can't seem to reconcile that with the stated result.

Can somebody help? What is that antiderivative anyway? I am quite lost here.

One interesting trick:

We have the Fourier transform pair:

$$ \int_{-\infty}^{t} x(\gamma) d\gamma <-> \frac{1}{j\omega} X(\omega) + \pi X(0) \delta(\omega) $$

Where $X(\cdot)$ is the FT pair of $x(\cdot)$.

Best Answer

We want to solve the following integral:

$$ I(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{jt(1+jt)} e^{\frac{-jNat}{1+jt}} e^{jxt} dt $$

To solve, we first recall a property of the Fourier transform. Namely:

$$ F^{-1}\big\{ \frac{G(t)}{jt} \big\} = \int_{-\infty}^{x} g(\tau) d\tau $$

where $g(x)$ is the inverse Fourier transform of $G(t)$, and $\int_{-\infty}^{\infty} g(\tau) d\tau = 0$. Thus we only need to solve:

$$ I_{1}(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{1+jt} e^{\frac{-jNat}{1+jt}} e^{jxt} dt $$

So, we start solving this integral by noticing:

$$ e^{\frac{-jNat}{1+jt}} = e^{-Na} e^{\frac{Na}{1+jt}} $$

Thus:

$$ I_{1}(x) = \frac{e^{-Na}}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{1+jt} e^{\frac{Na}{1+jt}} e^{jxt} dt $$

We use the power series of $e^{\frac{Na}{1+jt}}$ in the integral:

$$ I_{1}(x) = \frac{e^{-Na}}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{1+jt} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k!(1+jt)^{k}} e^{jxt} dt $$

Pulling out the constants and interchanging the sum and integral because the sum is convergent, we get:

$$ I_{1}(x) = \frac{e^{-Na}}{2 \pi} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k!} \int_{-\infty}^{\infty} \frac{1}{(1+jt)^{k+1}} e^{jxt} dt $$

The integral is the inverse Fourier transform of $\frac{1}{(1+jt)^{k+1}}$, which can be found either by the residue theorem using a contour in the upper half plane or by consulting Fourier tables. We get:

$$ I_{1}(x) = \frac{e^{-Na}}{2 \pi} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k!} \frac{2 \pi}{k!} x^{k} e^{-x} $$

Noting that:

$$I_{0}(x) = \sum_{k=0}^{\infty} \frac{\big(\frac{x}{2}\big)^{2k}}{k! \Gamma(k+1)}$$

Thus taking out the $e^{-x}$ and combining the powerts, we have:

$$ I_{1}(x) = e^{-x - Na} I_{0}(2 \sqrt{Nax})$$

So using the integration property of the Fourier transform mentioned earlier:

$$ I(x) = e^{-Na} \int_{0}^{x} e^{-y} I_{0}(2 \sqrt{Nay}) dy $$

since $I_{0}(x) = 0 \forall x <0$. This is the solution we were looking for.

Related Solutions

[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 $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$? .

Integrate $ I(x) = \frac{1}{2 \pi}\int\limits_{-\infty}^{\infty} \frac{1}{(1-jt)^{N}} e^{\frac{jaNt}{1-jt}} e^{-jxt} dt$

So I have found the solution.

Our problem is:

$$ I(x) = \frac{1}{2 \pi} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{(1-jt)^{N}} e^{\frac{jaNt}{1-jt}} e^{-jxt} dt $$

Where $j = \sqrt{-1}$, $a \in \mathbb{R}$ and $N \in \mathbb{Z_{++}}$.

Solution:

The first thing to do is negate $t$. This gives us:

$$ I(x) = \frac{1}{2 \pi} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{(1+jt)^{N}} e^{\frac{-jaNt}{1+jt}} e^{jxt} dt $$

Next, we notice that:

$$e^{\frac{-jaNt}{1+jt}} = e^{-Na} e^{\frac{Na}{1+jt}} $$

This leads us to:

$$ I(x) = \frac{1}{2 \pi} e^{-Na} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{(1+jt)^{N}} e^{\frac{Na}{1+jt}} e^{jxt} dt $$

Now we expand the first exponential into a power series:

$$e^{\frac{Na}{1+jt}} = \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k! (1+jt)^{k}} $$

Plugging this back in, we get:

$$ I(x) = \frac{1}{2 \pi} e^{-Na} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{(1+jt)^{N}} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k! (1+jt)^{k}} e^{jxt} dt $$

$$ I(x) = \frac{1}{2 \pi} e^{-Na} \displaystyle\int\limits_{-\infty}^{\infty} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k! (1+jt)^{N+k}} e^{jxt} dt $$

The sum is convergent, so we interchange the sum and integral:

$$ I(x) = \frac{1}{2 \pi} e^{-Na} \sum_{k=0}^{\infty} \displaystyle\int\limits_{-\infty}^{\infty} \frac{(Na)^{k}}{k! (1+jt)^{N+k}} e^{jxt} dt $$

We pull the constants out:

$$ I(x) = \frac{1}{2 \pi} e^{-Na} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k!} \displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{ (1+jt)^{N+k}} e^{jxt} dt $$

This integral is just the inverse Fourier transform of $\frac{1}{ (1+jt)^{N+k}}$. It can be computed using the residue theorem and taking a contour in the upper half plane -- or simply consulting Fourier tables. This gives us:

$$ I(x) = \frac{1}{2 \pi} e^{-Na} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k!} \frac{2 \pi}{(N+k-1)!} x^{N+k-1} e^{-x} $$

Cancelling the $2\pi$ and rearranging, we get:

$$ I(x) = e^{-x-Na} \sum_{k=0}^{\infty} \frac{(Na)^{k}}{k! (N+k-1)!} x^{N+k-1} $$

Now we notice that:

$$ x^{N+k-1} = x^{N-1} x^{k} $$

This gives:

$$ I(x) = e^{-x-Na} \sum_{k=0}^{\infty} x^{N-1} \frac{(Nax)^{k}}{k! (N+k-1)!} $$

Now we notice that:

$$ x^{N-1} = \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} (Nax)^{\frac{N-1}{2}} $$

Plugging this in:

$$ I(x) = \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{-x-Na} \sum_{k=0}^{\infty} \frac{(Nax)^{\frac{N-1}{2}} (Nax)^{k}}{k! (N+k-1)!} $$

Combing powers:

$$ I(x) = \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{-x-Na} \sum_{k=0}^{\infty} \frac{\sqrt{Nax}^{N-1+2k}}{k! (N+k-1)!} $$

Thus:

$$ I(x) = \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{-x-Na} I_{N-1}( 2\sqrt{Nax} ) $$

As desired.

Best Answer

Related Solutions

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

Integrate $ I(x) = \frac{1}{2 \pi}\int\limits_{-\infty}^{\infty} \frac{1}{(1-jt)^{N}} e^{\frac{jaNt}{1-jt}} e^{-jxt} dt$

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