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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?

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    $\begingroup$ An equivalent form of you integral: $\displaystyle I=\frac12\int_0^\infty J_0(x)\operatorname{arccot} x\,dx$. $\endgroup$ – Vladimir Reshetnikov Nov 22 '13 at 18:59
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    $\begingroup$ And another closed form (although not as simple and nice as in Ron Gordon's answer): $\displaystyle I=\frac\pi4-\frac18G_{1,3}^{3,0}\left(\tfrac14\middle|\begin{array}c\tfrac12\\ - \tfrac12,0,0\\\end{array}\right)$. $\endgroup$ – Vladimir Reshetnikov Nov 22 '13 at 19:11
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    $\begingroup$ @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) $\endgroup$ – Ron Gordon Nov 25 '13 at 14:51
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    $\begingroup$ (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": math.stackexchange.com/questions/562769/…) I think this should have satisfied the OP's reasonable request. $\endgroup$ – Ron Gordon Nov 25 '13 at 14:54
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    $\begingroup$ @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. $\endgroup$ – Vladimir Reshetnikov Nov 25 '13 at 20:41
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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.

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  • $\begingroup$ @VladimirReshetnikov: thanks, that's really kind of you. $\endgroup$ – Ron Gordon Nov 22 '13 at 18:47
  • $\begingroup$ @ Ron Gordon: Take a look at my timing 1.014 s when it is calculated numerically . $\endgroup$ – user64494 Nov 22 '13 at 19:30
  • $\begingroup$ @ Ron Gordon: This is an old-fashioned math, isn't it? $\endgroup$ – user64494 Nov 22 '13 at 19:33
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    $\begingroup$ @user64494: not sure what you mean by "old-fashioned math." But i can say that, even at 1 sec evaluation time, if we had to evaluate many thousands of values (say, $-p x$ rather than $x$), then finding the analytical result is even more valuable. $\endgroup$ – Ron Gordon Nov 22 '13 at 19:36
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    $\begingroup$ @cleo: words fail. Thank you. $\endgroup$ – Ron Gordon Nov 26 '13 at 18:42

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