Integral $\int_0^\infty{_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)\frac{dx}{1+4\,x}$ I need to evaluate this integral to a high precision:
$$\large I=\int_0^\infty{_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)\frac{dx}{1+4\,x}$$
Symbolic integration in Mathematica cannot handle this integral. When I try to evaluate it numerically, it converges extremely slowly and the result is very unstable, so I am not even sure how many correct digits I got. It looks kinda $I\stackrel?\approx0.6212...$
So, my only hope is to find a closed form for $I$ in terms of functions for which fast and robust numerical algorithms exist. Could you please help me to find it? 
 A: 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.
