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

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 $$N-1$$

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?

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....

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?

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:

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

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

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.

• In case that matters for your application, note that $I(x) = 0$ for $x < 0$ and $I(x) = (x/N)^{(N - 1)/2} (a^{-1/2})^{N - 1} e^{-x - N a} I_{N - 1}(2 \sqrt {N a x})$ for $x > 0 \land a < 0$. Commented Aug 1, 2019 at 17:05
• @Maxim Yea so if $a = 0$, it is undefined, right? So but what if $a > 0$? Shouldn't that work, or no? Commented Aug 1, 2019 at 17:12
• Your result is correct for $x > 0 \land a > 0$. You just need to consider that you have to take lower semicircles instead of upper semicircles if $x < 0$ and that $(a^{-1/2})^{N - 1} \neq (1/a)^{(N - 1)/2}$ for $a < 0$ if $N$ is even. For $a = 0$, $I(x)$ can be computed as the limit: $\lim_{a \to 0} I(x) = x^{N - 1} e^{-x}/\Gamma(N)$. Commented Aug 1, 2019 at 17:36
• @clathratus What do you mean? Commented Aug 5, 2019 at 15:38