Fourier Transform of $f(x) = e^{-(x + a)} I_{0} ( 2 \sqrt{a x} )$

Question

So I am trying to solve the following integral:

$$ I(t) = \displaystyle\int\limits_{0}^{\infty} e^{-(x+a)} I_{0}(2 \sqrt{ax}) e^{jxt} dx $$

Where $j = \sqrt{-1}$ and $I_{0}(\cdot)$ is the modified Bessel function of the 1st kind of order 0. I know the solution is:

$$ I(t) = \frac{1}{1-jt} e^{\frac{jat}{1-jt}} $$

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My attempt:

We know that:

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

Therefore:

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

So we have:

$$ I(t) = \int_{0}^{\infty} e^{-(x+a)} \sum_{k=0}^{\infty} \frac{a^{k} x^{k}}{k! \Gamma(k+1)} e^{jxt} dx $$

$$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k}}{k! \Gamma(k+1)} \int_{0}^{\infty} x^{k} e^{-(x+a) + jxt} dx $$

$$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k} e^{-a}}{k! \Gamma(k+1)} \int_{0}^{\infty} x^{k} e^{-(1-jt)x} dx $$

The integral is the Laplace transform of $x^{k}$ at $s = (1-jt)$. Thus:

$$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k} e^{-a}}{\Gamma(k+1) (1-jt)^{k-1}} $$

However this is nowhere near the solution I am searching for. I am not really sure how to work with $I_{0}(\cdot)$ in integrals.

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How can I get the desired solution?

Answer 1

This solution is thanks to @metamorphy's tips.

We want to solve:

$$ I(t) = \int_{0}^{\infty} e^{-(x+a)} I_{0}(2 \sqrt{ax}) e^{jxt} dx $$

To start, we know that:

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

Therefore:

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

So then we want to solve:

$$ I(t) = \int_{0}^{\infty} e^{-(x+a)} \sum_{k=0}^{\infty} \frac{a^{k}x^{k}}{k! \Gamma(k+1)} e^{jxt} dx $$

Since the sum is convergent, we have:

$$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k}}{k! \Gamma(k+1)} \int_{0}^{\infty} x^{k} e^{-(x+a)} e^{jxt} dx $$ $$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k}}{k! \Gamma(k+1)} \int_{0}^{\infty} x^{k} e^{-(x+a) + jxt} dx $$ $$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k}e^{-a}}{k! \Gamma(k+1)} \int_{0}^{\infty} x^{k} e^{-(1-jt)x} dx $$

We note that the integral is simply the Laplace Transofmr of $x^{k}$ at $s = 1-jt$. Thus we get:

$$ I(t) = \sum_{k=0}^{\infty} \frac{a^{k} e^{-a}}{k! \Gamma(k+1)} \frac{k!}{(1-jt)^{k-1}}$$ $$ I(t) = e^{-a} \sum_{k=0}^{\infty} \frac{a^{k} }{\Gamma(k+1) (1-jt)^{k-1}}$$

Since $k \in \mathbb{Z}$, we know $\Gamma(k+1) = k!$, thus we get:

$$ I(t) = e^{-a} \sum_{k=0}^{\infty} \frac{1}{(1-jt)^{k-1}} \frac{a^{k} }{k! }$$

We recall the power series of $e^{x}$:

$$ e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!} $$

So we rearrange:

$$ I(t) = e^{-a} \sum_{k=0}^{\infty} \frac{\big(\frac{a}{(1-jt)}\big)^{k} }{k! (1-jt)} = \frac{e^{-a}}{(1-jt)} \sum_{k=0}^{\infty} \frac{\big(\frac{a}{(1-jt)}\big)^{k} }{k! } = \frac{e^{-a}}{(1-jt)} e^{\frac{a}{1-jt}} $$

Combining exponents we get: $$ I(t) = \frac{1}{1-jt} e^{\frac{a}{1-jt} - \frac{(1-jt)a}{1-jt}} = \frac{1}{1-jt} e^{\frac{a-a + jat}{1-jt}} = \frac{1}{1-jt} e^{\frac{jat}{1-jt}} $$

Thus we have come to the final solution:

$$ I(t) = \frac{1}{1-jt} e^{\frac{jat}{1-jt}} $$

Answer 2

An alternate method using the Laplace transform:

Let $u = ax$ and $s = (1-it)/a$. Then this integral is $$ I= \frac{e^{-a}}{a}\int_0^\infty I_0(2\sqrt{u})e^{-su}du $$ i.e. the Laplace transform of $I_0(2\sqrt{u})$. Since $I_0(2\sqrt{u})$ satisfies the differential equation $z f''+f'-f = 0$, its Laplace transform satisfies the differential equation $s^2F'(s) +(s+1)F(s)=0$, the solution of which is $F(s) = \exp(1/s)/s$. So $$ I = \frac{e^{-a}}{a}\frac{e^{1/s}}{s} = \frac{e^{-a}e^{\frac{a}{1-it}}}{1-it}= \frac{e^{\frac{ait}{1-it}}}{1-it} $$