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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?

Oksana Gimmel
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    An equivalent form of you integral: $\displaystyle I=\frac12\int_0^\infty J_0(x)\operatorname{arccot} x,dx$. – Vladimir Reshetnikov Nov 22 '13 at 18:59
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    And another closed form (although not as simple and nice as in Ron Gordon's answer): $\displaystyle I=\frac\pi4-\frac18G_{1,3}^{3,0}\left(\tfrac14\middle|\begin{array}c\tfrac12\ - \tfrac12,0,0\\end{array}\right)$. – Vladimir Reshetnikov Nov 22 '13 at 19:11
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    Why do you need to compute this to a high precision? This question lacks crucial context such as where you encountered the problem and what you have tried already. – Carl Mummert Nov 25 '13 at 04:03
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    @CarlMummert: the ocntext (which I have observed myself) is that Mathematica takes an unusually long time to evaluate this integral numerically. At least with v. 8.0.4, one can numerically integrate improper integrals such as those above, but in this case, over two hours of computation on a current processor yielded nothing. Truncating the integration interval yielded results, but the convergence was maddeningly slow. That alone should be sufficient motivation to ask if there is a closed form that may be evaluated much more quickly. (1/2) – Ron Gordon Nov 25 '13 at 14:51
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    (2/2) Turns out that, indeed this is the case. Fast algorithms exists for even the modified Struve function, so that the time to evaluate the closed form was quite negligible. (Thus, satisfying my definition of "closed form": http://math.stackexchange.com/questions/562769/what-would-qualify-as-a-valid-reason-to-believe-there-is-a-closed-form/562810#562810) I think this should have satisfied the OP's reasonable request. – Ron Gordon Nov 25 '13 at 14:54
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    @Ron Gordon: I would be more interested to hear where the integral came from. Surely there are many integrals that also cannot be integrated symbolically in Mathematica. If you have also observed this yourself, where did you get the integral from? – Carl Mummert Nov 25 '13 at 14:57
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    @CarlMummert: of course there are many integrals that Mathematica cannot do. There are, however, somewhat fewer integrals that Mathematica cannot do, yet we inferior humans still can. This is apparently one of them. (Although, to be fair, the integral was posed in a form with which Mathematica is not yet equipped to handle well. Maple neither apparently. These things can change of course.) I do not understand your question, though: I have no idea how this integral arose; only the OP can answer that. – Ron Gordon Nov 25 '13 at 15:03
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    @CarlMummert My personal feeling is that always asking users to explain context would place an unnecessary burden on them. A certain integral could came up in many-pages-long chain of computations that originated in some physical or economical model. I do not see how it would be useful in general to post the full context here. Many integrals are interesting self-contained problems that could be used to demonstrate some ingenious approaches and techniques that could be widely applicable irrespectively of where the integral came from. – Vladimir Reshetnikov Nov 25 '13 at 20:41

1 Answers1

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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.

Ron Gordon
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