I am looking for a closed form for these two integrals
$$\int_{-\infty}^{-a}\text{d}x \frac{1}{|x|}e^{-\frac{1}{2}x^2\sigma^2}e^{i k |x|}+\int_a^{\infty}\text{d}x \frac{1}{|x|}e^{-\frac{1}{2}x^2\sigma^2}e^{i k |x|}$$ $$\int_{0}^{\infty}\text{d}xJ_0(k\,x)e^{-\frac{1}{2}x^2\sigma^2}e^{-i k x}$$
where $a>0$, $\sigma>0$ and $k\geq0$, and $J_0(x)$ is the Bessel function of first kind.
Is it possible to find a closed form for them? Any help with how to obtain the expressions or simply the final answer would be highly appreciated. Even insight for the indefinite integrals would be useful.
EDIT: thanks to @Yuriy's advice, the second integral is
$$\int_{0}^{\infty}\text{d}xJ_0(k\,x)e^{-\frac{1}{2}x^2\sigma^2}e^{-i k x}=\frac{1}{2\sigma ^2}\left[\sqrt{2 \pi } \sigma \, _2F_2\left(\frac{1}{4},\frac{3}{4};\frac{1}{2},1;-\frac{2 k^2}{\sigma ^2}\right)-2 i k \, _2F_2\left(\frac{3}{4},\frac{5}{4};\frac{3}{2},\frac{3}{2};-\frac{2 k^2}{\sigma ^2}\right)\right]$$
The first sum of integrals is really just one integral (the function is even):
$$\int_{-\infty}^{-a}\text{d}x \frac{1}{|x|}e^{-\frac{1}{2}x^2\sigma^2}e^{i k |x|}+\int_a^{\infty}\text{d}x \frac{1}{|x|}e^{-\frac{1}{2}x^2\sigma^2}e^{i k |x|}=2\int_a^{\infty}\text{d}x \frac{1}{|x|}e^{-\frac{1}{2}x^2\sigma^2}e^{i k |x|}$$
Now, since $a>0$, we can get rid of the absolute value function:
$$I_1(k,\sigma,a)=2\int_a^{\infty}\text{d}x \frac{1}{x}e^{-\frac{1}{2}x^2\sigma^2}e^{i k x}$$
$$\frac{\partial I_1(k,\sigma,a)}{\partial k}=2 i \int_a^{\infty} e^{-\frac{1}{2}x^2\sigma^2}e^{i k x} \text{d}x$$
This integral can be expressed as error function of imaginary variable. I'm not sure how to work with it, so I'm leaving the rest up to you.
Now for the second integral.
$$I_2(k, \sigma)=\int_{0}^{\infty}\text{d}xJ_0(k\,x)e^{-\frac{1}{2}x^2\sigma^2}e^{-i k x}$$
First, let's change the variable and parameter to simplify the expression $kx \to x$ and $\sigma / (\sqrt{2}~k)=\sigma$:
$$I_2(k, \sigma)=\frac{1}{k} \int_{0}^{\infty}J_0(x)e^{-x^2\sigma^2}e^{-i x}\text{d}x$$
Then the one way is to use the integral representation of the Bessel function:
$$J_0(x)=\frac{1}{\pi} \int_0^{\pi} \cos (x \cos t) dt=\frac{1}{2\pi} \left( \int_0^{\pi} e^{i x \cos t} dt+\int_0^{\pi} e^{-i x \cos t} dt\right)$$
It's enough to consider one of two terms, another can be treated in the same way, then we get in our original integral:
$$\frac{1}{2\pi k} \int_0^{\pi} \int_{0}^{\infty}e^{-x^2\sigma^2}e^{-i x (1+\cos t)}\text{d}x ~\text{d}t$$
Now we complete the square:
$$-\sigma^2 x^2-i (1+\cos t)x=-\sigma^2 \left(x+i \frac{1+\cos t}{2} \right)^2-\sigma^2 \frac{(1+\cos t)^2}{4}$$
However, this way we get error function of imaginary variable under the integral multiplied by the exponent with $-(1+\cos t)^2$ as an argument. So I think this just makes it worse. If we had $-\infty$ as a second limit the integral could be expressed as a series of modified Bessel functions, but in your case I don't see any way to simplify it.