Integral $\int_0^\infty{_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)\frac{dx}{1+4\,x}$

OK, I have an analytical result:

$$\frac{\pi}{4} \left [K_0(1) \mathbf{L}_{-1}(1) + K_1(1) \mathbf{L}_{0}(1)\right ] \approx 0.621255$$

where $K_0$ and $K_1$ are modified Bessel functions of the second kind, and $\mathbf{L}_{0}$ and $\mathbf{L}_{-1}$ are modified Struve functions of the first kind.

This may be obtained by recognizing that (+)

$${_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right) = \int_0^1 du \, J_0\left ( 2 u \sqrt{x}\right ) $$

The integral is then, upon reversing order,

$$\int_0^1 du \, \int_0^{\infty} dx \frac{J_0\left ( 2 u \sqrt{x}\right )}{1+4 x}$$

The following will need a derivation (++):

$$\int_0^{\infty} dx \frac{J_0\left ( 2 u \sqrt{x}\right )}{1+4 x} = \frac12 K_0(u)$$

The stated result is then

$$\frac12 \int_0^1 du \, K_0(u) = \frac{\pi}{4} \left [K_0(1) \mathbf{L}_{-1}(1) + K_1(1) \mathbf{L}_{0}(1)\right ] $$

which may be found in the DLMF.

Derivation of (+)

Note that the coefficient of $x^n$ in ${_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)$, by definition, is

$$a_n = \frac{\frac12 \left (\frac12 + 1 \right ) \left (\frac12 + 2 \right ) \cdots \left (\frac12 + n-1 \right )}{n! \frac{3}{2} \left (\frac{3}{2} + 1 \right ) \left (\frac{3}{2} + 2 \right ) \cdots \left (\frac{3}{2} + n-1 \right )} \frac{(-1)^n}{n!}$$

which, after simplification, is

$$a_n = \frac{(-1)^n}{(2 n+1) (n!)^2} $$

Then, note that

$$J_0(2 u \sqrt{x}) = \sum_{n=0}^{\infty} (-1)^n \frac{u^{2 n}}{(n!)^2} x^n$$

Then

$$\int_0^1 du \, J_0(2 u \sqrt{x}) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)(n!)^2} x^n$$

and one may see that the coefficients of the respective power series are equal.

Derivation of (++)

Sub $x=r^2$ to get

$$\int_0^{\infty} dr \, r \frac{J_0(2 u r)}{1+4 r^2}$$

Now write

$$J_0(2 u r) = \frac{1}{2 \pi} \int_0^{2 \pi} d\theta \, e^{i 2 u r \cos{\theta}}$$

so that we now have as the integral

$$\frac{1}{2 \pi} \int_0^{2 \pi} d\theta \, \int_0^{\infty} dr \, r \frac{e^{i 2 u r \cos{\theta}}}{1+4 r^2}$$

Note that we may simply convert back to rectangular coordinates to get

$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} dx \, e^{i 2 u x} \, \int_{-\infty}^{\infty} \frac{dy}{1+4 x^2+4 y^2} $$

The inner integral is simple, so we are back to a single integral:

$$\frac18 \int_{-\infty}^{\infty} dx \frac{e^{i 2 u x}}{\sqrt{1+4 x^2}} $$

By subbing $x=\frac12 \sinh{t}$ and using the definition

$$K_0(u) = \int_0^{\infty} dt \, \cos{(u \, \sinh{t} )}$$

we obtain the stated result.