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

Question

I would like to solve the following integral:

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

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

The answer I am looking for is:

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

Where $I_{N-1}(\cdot)$ is the modified Bessel function of the first kind of order $N-1$

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My approach:

First we note that we can change the sign of $t$ because the limits are symmetric. This gives us:

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

I want to do this because I want to rewrite $(1-jt)^{N}$. Namely, we write:

$$ (1+jt)^{N} = j^{N}(t-j)^{N} $$

This gives us:

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

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

Now we remember that $I_{N}(x) = \sum_{k=0}^{\infty} \frac{\big(\frac{x}{2}\big)^{N+2k}}{k! \Gamma(N+k+1)} $. So to get closer to this, I expand the exponential $e^{\frac{-aNt}{t-j}}$ into its power series expansion:

$$ e^{\frac{-aNt}{t-j}} = \sum_{k=0}^{\infty} \frac{(-aNt)^{k}}{k!(t-j)^{k}} $$

So plugging this in, and interchanging the sum and integral gives us:

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

Now we form a half circular contour in the upper half plane. This gives a pole at $z=j$. So we can apply the residue thm. So we have:

$$ I(x) = \frac{1}{j^{N} 2 \pi} \sum_{k=0}^{\infty} \frac{1}{k!} \displaystyle\int\limits_{C} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} dz = \lim_{R \rightarrow \infty} \Big (\frac{1}{j^{N} 2 \pi} \sum_{k=0}^{\infty} \frac{1}{k!} \displaystyle\int\limits_{C_{R}} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} dz + \frac{1}{j^{N} 2 \pi} \sum_{k=0}^{\infty} \frac{1}{k!} \displaystyle\int\limits_{-R}^{R} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} dz \Big) $$

Clearly as $R \rightarrow \infty$, $| \int_{C_{R}} \cdot | \rightarrow 0$ So we have:

$$ I(x) = \frac{1}{j^{N} 2 \pi} \sum_{k=0}^{\infty} \frac{1}{k!} \displaystyle\int\limits_{C} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} dz = \lim_{R \rightarrow \infty} \frac{1}{j^{N} 2 \pi} \sum_{k=0}^{\infty} \frac{1}{k!} \displaystyle\int\limits_{-R}^{R} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} dz $$

So the we want to solve $| \int_{C} |$, which is:

$$ \frac{1}{j^{N} 2 \pi} \sum_{k=0}^{\infty} \frac{1}{k!} \displaystyle\int\limits_{C} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} dz = j 2 \pi Res\{z=j\} $$

The reisude is then:

$$Res\{z=j\} = \frac{1}{(N+k-1)!} lim_{z \rightarrow j} \frac{d^{N+k-1}}{dz^{N+k-1}} (z-j)^{N+k} \frac{(-aNz)^{k}}{(z-j)^{N+k}} e^{jxz} = \frac{1}{(N+k-1)!} \frac{d^{N+k-1}}{dz^{N+k-1}} (-aNz)^{k} e^{jxz} $$

This is where I get stuck. I use the generalized Leibniz rule to compute the residue, as it is an $N+k$ order pole. However, what I get is nothing close to the solution I am searching for.

Is this the right approach? What am I doing wrong? How can I solve this integral?

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Progress:

So I noticed that we can rewrite the one exponential. We get:

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

This leads us to the following integral:

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

In other words, the $e^{-Na}$ is out, like in the solution. Also, note that $lim_{z \rightarrow j} e^{jxt} = e^{-x}$ And $ e^{-Na} e^{-x} = e^{-x - Na}$. So it seems we need to keep the $e^{jxt}$ term until we take the limit of the residue?

I am still confused as to where the $\Big(\frac{x}{Na}\Big)^\frac{{N-1}}{2}$ comes from....

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Another hint:

We notice that:

$$ I_{N-1}(2 \sqrt{Nax}) = \sum_{k=0}^{\infty} \frac{(Nax)^{\frac{N-1}{2} (Nax)^{k}}}{k! (N+k-1)!} $$

Also that:

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

So maybe we need to get $x^{N-1}$ out somehow. Where does the $(Nax)$ term come from though?

Answer 1

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 $N \in \mathbb{Z_{++}}$.

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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.