In my research involving quantum field theory and computing correlation functions in perturbation theory, I encountered the following integral that seems pretty hard to solve:
$$f(x)=\int_{0}^{\infty}\frac{dz}{z(z^2+1)}\text{arcsinh}^2(z)J_0(xz)$$
I wonder if anyone has any insights towards a solution. A series expansion and it's convergence properties would be fantastic as well.
On
Not elegant at all since using brute force.
Computing numerically the integrals, it seems that a good (totally empirical) approximation could be $$\log(f(x))=a-b\, x^c$$ Based of $40$ equally spaced data points between $0.5$ and $20.0$, a quick and dirty nolinear regression gives $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.17160 & 0.00218 & \{0.17603,0.16717\} \\ b & 1.10970 & 0.00091 & \{1.10785,1.11155\} \\ c & 0.97817 & 0.00025 & \{0.97765,0.97868\} \\ \end{array}$$
Consider the contour representation $$arcsinh^2(z)=\int\limits_{c-i\infty }^{c+i\infty }{\frac{{{\pi }^{3/2}}\Gamma \left( -\tfrac{1}{2}s \right){{z}^{-s}}}{2s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)\sin \left( \tfrac{1}{2}\pi s \right)}ds}$$ where we choose $-1<c<0$ so the contour encircles all poles on the negative axis. We have then $$f\left( x \right)=\int\limits_{0}^{\infty }{\frac{{{z}^{-s-1}}{{J}_{0}}\left( xz \right)}{{{z}^{2}}+1}dz}\int\limits_{c-i\infty }^{c+i\infty }{\frac{{{\pi }^{3/2}}\Gamma \left( -\tfrac{1}{2}s \right)}{2s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)\sin \left( \tfrac{1}{2}\pi s \right)}ds}$$ Now things will get somewhat murky. For large x we can approximate the Bessel function to obtain $$f\left( x \right)\simeq \sqrt{\frac{2}{\pi x}}\int\limits_{0}^{\infty }{\frac{{{z}^{-s-3/2}}\cos \left( xz-\tfrac{1}{4}\pi \right)}{{{z}^{2}}+1}dz}\int\limits_{c-i\infty }^{c+i\infty }{\frac{{{\pi }^{3/2}}\Gamma \left( -\tfrac{1}{2}s \right)}{2s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)\sin \left( \tfrac{1}{2}\pi s \right)}ds}.$$
Now the integral $$\int\limits_{0}^{\infty }{\frac{{{z}^{-s-3/2}}\cos \left( xz-\tfrac{1}{4}\pi \right)}{{{z}^{2}}+1}dz}=\frac{1}{\sqrt{2}}\int\limits_{0}^{\infty }{\frac{{{z}^{-s-3/2}}\cos \left( xz \right)}{{{z}^{2}}+1}dz}+\frac{1}{\sqrt{2}}\int\limits_{0}^{\infty }{\frac{{{z}^{-s-3/2}}\sin \left( xz \right)}{{{z}^{2}}+1}dz},$$ is listed in tables and has an expression via hypergeometric functions (see GR 3.766 onwards for example). I’ll omit a lot of algebra here, but you can boil it down to something like the following $$f\left( x \right)=\int\limits_{c-i\infty }^{c+i\infty }{\frac{\sqrt{2}{{x}^{5/2+s}}\Gamma \left( -\tfrac{5}{2}-s \right)\Gamma \left( -\tfrac{1}{2}s \right)}{s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)}F\left[ s \right]ds}+\frac{{{\pi }^{2}}}{\sqrt{2x}}\int\limits_{c-i\infty }^{c+i\infty }{\frac{\Gamma \left( 1-\tfrac{1}{2}s \right)\left( \cosh \left( x \right)\left( \cot \left( \tfrac{1}{2}\pi \right)-1 \right)-\sinh \left( x \right)\left( \cot \left( \tfrac{1}{2}\pi \right)+1 \right) \right)}{{{s}^{2}}\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)\cos \left( \pi s \right)}ds}$$ where $-1<c<-1/2$ and where $F\left[ s \right]$ is a hypergeometric function that has no poles on the negative real axis. Accounting for poles at odd-integer multiples of $1/2$, then the terms generated by the residues of the gamma function multiplying the hypergeometric function, cancel several terms from the residues at odd-integer halves from the second contour integral. So calculating these residues and cancelling terms, we find then
$$f\left( x \right)\simeq \frac{{{\pi }^{5/2}}{{e}^{-x}}}{\sqrt{32x}}-\frac{8\pi }{\sqrt{2x}}\sum\limits_{n=2}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\Gamma \left( \tfrac{3}{2}+n \right)}{{{\left( 2n+1 \right)}^{2}}\Gamma \left( \tfrac{3}{4}+\tfrac{1}{2}n \right)}\frac{\left( \cosh \left( x \right)\sin \left( \tfrac{1}{2}n\pi \right)+\sinh \left( x \right)\cos \left( \tfrac{1}{2}n\pi \right) \right)}{\cos \left( \tfrac{1}{2}n\pi \right)+\sin \left( \tfrac{1}{2}n\pi \right)}}$$ The series here is divergent in n (and the terms also diverge for large x), and including additional terms doesn't improve the approximation. So dropping those terms we have a leading term approximation $$f\left( x \right)\simeq \frac{{{\pi }^{5/2}}{{e}^{-x}}}{\sqrt{32x}}$$
The justification for all of this obviously needs a lot more work and my guess is using more terms in the approximation of the Bessel function might clean up the divergent series (in x). i also think a much more careful examination of convergence along the contours would result in choosing a contour that yields a 'remainder' series that is truly asymptotic in x, and where additional terms improve the approximation. However as it stands i think this very rough sketch is indeed picking up the leading asymptotic behaviour of the integral. Below is a plot of the numerical evaluation of the integral (solid) and the asymptotic approximation (dashed) (note the domain of the plot is purposefully avoiding the singularity in the asymptotic approximation - it is for large x, but starts to become quite reasonable for x>5). I can't say for certain that adding more terms from the approximation of the Bessel function would improve the result.
Rather than approximate the Bessel function let’s return to the following: $$f\left( x \right)=\int\limits_{0}^{\infty }{\frac{{{z}^{-s-1}}{{J}_{0}}\left( xz \right)}{{{z}^{2}}+1}dz}\int\limits_{c-i\infty }^{c+i\infty }{\frac{{{\pi }^{3/2}}\Gamma \left( -\tfrac{1}{2}s \right)}{2s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)\sin \left( \tfrac{1}{2}\pi s \right)}ds}$$ There is a form of this in GR, and again involves hypergeometric functions, however mathematica yields the following $$f\left( x \right)=\int\limits_{c-i\infty }^{c+i\infty }{\frac{{{\pi }^{5/2}}\Gamma \left( -\tfrac{1}{2}s \right)\left( {{x}^{2+s}}F\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right)-{{2}^{2+s}}{{I}_{0}}\left( x \right) \right)}{{{2}^{s+4}}s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right){{\sin }^{2}}\left( \tfrac{1}{2}\pi s \right)}ds}$$ Where ${{I}_{0}}\left( x \right)$ is the modified Bessel function and the F is the regularised hypergeometric function. The only poles on the negative axis are those from the sine function which are order two. We don’t have the issue as we had before where the approximation introduced poles at odd-half integers. Note $$\frac{1}{{{\sin }^{2}}\left( \tfrac{1}{2}\pi z \right)}\underset{z\to -2n}{\mathop{=}}\,\left( \frac{4}{{{\pi }^{2}}{{\left( z+2n \right)}^{2}}}+\frac{1}{3}+\frac{{{\pi }^{2}}}{60}{{\left( z+2n \right)}^{2}}+\frac{2{{\pi }^{4}}}{1512}{{\left( z+2n \right)}^{4}} \right)$$ And if $f\left( s \right)=f\left( -2n \right)+f'\left( -2n \right)\left( s+2n \right)+\frac{1}{2}f''\left( -2n \right){{\left( s+2n \right)}^{2}}$ then $$\frac{f\left( z \right)}{{{\sin }^{2}}\left( \tfrac{1}{2}\pi z \right)}\underset{z\to -2n}{\mathop{=}}\,\frac{4f'\left( -2n \right)}{{{\pi }^{2}}\left( z+2n \right)}+O\left( 1 \right)$$ Hence $$\begin{align} & \frac{{{\pi }^{5/2}}\Gamma \left( -\tfrac{1}{2}s \right)\left( {{x}^{2+s}}F\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right)-{{2}^{2+s}}{{I}_{0}}\left( x \right) \right)}{{{2}^{s+4}}s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right){{\sin }^{2}}\left( \tfrac{1}{2}\pi s \right)} \\ & \underset{s\to -2n}{\mathop{=}}\,\frac{{{2}^{-3-s}}\sqrt{\pi }\Gamma \left( -\tfrac{1}{2}s \right)}{{{s}^{2}}\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)\left( z+2n \right)}\left\{ -{{2}^{2+s}}{{I}_{0}}\left( x \right)\left( -2+s\psi \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)-s\psi \left( -\tfrac{1}{2}s \right) \right)-{{x}^{2+s}}F\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right)\left( 2-s\log \left( \tfrac{1}{4}{{x}^{2}} \right)-s\psi \left( \tfrac{1}{2}-\tfrac{1}{2}s \right)+s\psi \left( -\tfrac{1}{2}s \right) \right)+s{{x}^{2+s}}{{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right)+s{{x}^{2+s}}{{F}^{\left( \left\{ 0 \right\},\left\{ 0,1 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right) \right\}+O\left( 1 \right) \\ \end{align}$$ Therefore, the residues are $$\begin{align} & \underset{s=-2n}{\mathop{res}}\,\frac{{{\pi }^{5/2}}\Gamma \left( -\tfrac{1}{2}s \right)\left( {{x}^{2+s}}F\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right)-{{2}^{2+s}}{{I}_{0}}\left( x \right) \right)}{{{2}^{s+4}}s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right){{\sin }^{2}}\left( \tfrac{1}{2}\pi s \right)} \\ & =\frac{{{2}^{-3+2n}}{{2}^{2-2n}}\sqrt{\pi }\Gamma \left( n \right)}{{{\left( 2n \right)}^{2}}\Gamma \left( \tfrac{1}{2}+n \right)}\left\{ {{I}_{0}}\left( x \right)\left( 2+2n\psi \left( \tfrac{1}{2}+n \right)-2n\psi \left( n \right) \right)-{{\left( \frac{x}{2} \right)}^{2-2n}}F\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right)\left( 2+2n\log \left( \tfrac{1}{4}{{x}^{2}} \right)+2n\psi \left( \tfrac{1}{2}+n \right)-2n\psi \left( n \right) \right)-2n{{\left( \frac{x}{2} \right)}^{2-2n}}\left( {{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right)+{{F}^{\left( \left\{ 0 \right\},\left\{ 0,1 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right) \right) \right\} \\ \end{align}$$ Upon noting $$F\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right)=\frac{{{x}^{2\left( n-1 \right)}}}{{{4}^{n-1}}}{}_{0}{{F}_{1}}\left( 1,\tfrac{1}{4}{{x}^{2}} \right)={{\left( \frac{x}{2} \right)}^{2n-2}}{{I}_{0}}\left( x \right)$$ and that the two derivatives of the hypergeometric function yield the same function, we can cancel and simplify terms to obtain $$\underset{s=-2n}{\mathop{res}}\,\frac{{{\pi }^{5/2}}\Gamma \left( -\tfrac{1}{2}s \right)\left( {{x}^{2+s}}F\left( \left\{ 1 \right\},\left\{ 2+\tfrac{1}{2}s,2+\tfrac{1}{2}s \right\},\tfrac{1}{4}{{x}^{2}} \right)-{{2}^{2+s}}{{I}_{0}}\left( x \right) \right)}{{{2}^{s+4}}s\Gamma \left( \tfrac{1}{2}-\tfrac{1}{2}s \right){{\sin }^{2}}\left( \tfrac{1}{2}\pi s \right)}=\frac{\sqrt{\pi }\Gamma \left( n \right)}{4n\Gamma \left( \tfrac{1}{2}+n \right)}\left\{ -{{I}_{0}}\left( x \right)\log \left( \tfrac{1}{4}{{x}^{2}} \right)-2{{\left( \frac{x}{2} \right)}^{2-2n}}{{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right) \right\}$$ resulting in $$f\left( x \right)=-\sum\limits_{n=1}^{\infty }{\frac{\sqrt{\pi }\Gamma \left( n \right)}{4n\Gamma \left( \tfrac{1}{2}+n \right)}\left\{ \log \left( \tfrac{1}{4}{{x}^{2}} \right){{I}_{0}}\left( x \right)+2{{\left( \frac{x}{2} \right)}^{2-2n}}{{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right) \right\}}$$ This appears to be an asymptotic series valid for large x (and the reason for this will have something to do with the choice of contours in the Mellin transform). The advantage of this approximation over the last is that we avoid a divergent series, however gain two functions, each diverging as x becomes large, but together cancel each other out creating a function $\to 0$ as $x\to \infty $, which I think is true no matter how many terms we take because every term involving the hypergeometric function essentially behaves like the last. That is it appears the term involving the hypergeometric function is independent of n. To see why this might be observe firstly $$F\left( \left\{ a \right\},\left\{ b,c \right\},z \right)=\frac{1}{\Gamma \left( b \right)\Gamma \left( c \right)}{}_{1}{{F}_{2}}\left( \left\{ a \right\},\left\{ b,c \right\},z \right)=\frac{1}{\Gamma \left( a \right)}\sum\limits_{k=0}^{\infty }{\frac{\Gamma \left( a+k \right)}{\Gamma \left( b+k \right)\Gamma \left( c+k \right)}\frac{{{z}^{k}}}{k!}}$$ Hence we have something like $$2{{\left( \frac{x}{2} \right)}^{2-2n}}{{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right)=-4{{\left( \frac{x}{2} \right)}^{2-2n}}\sum\limits_{k=0}^{\infty }{\frac{\psi \left( 2-n+k \right)}{\Gamma {{\left( 2-n+k \right)}^{2}}}{{\left( \frac{x}{2} \right)}^{2k}}}$$ But all terms in the series are zero until $k=n-1$ and so we have $$2{{\left( \frac{x}{2} \right)}^{2-2n}}{{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 2-n,2-n \right\},\tfrac{1}{4}{{x}^{2}} \right)=-4{{\left( \frac{x}{2} \right)}^{2-2n}}\sum\limits_{k=0}^{\infty }{\frac{\psi \left( 2-n+k \right)}{\Gamma {{\left( 2-n+k \right)}^{2}}}{{\left( \frac{x}{2} \right)}^{2k}}}=-4\sum\limits_{m=0}^{\infty }{\frac{\psi \left( 1+m \right)}{\Gamma {{\left( 1+m \right)}^{2}}}{{\left( \frac{x}{2} \right)}^{2m}}}$$ Which is independent of n. However the above series is not what is being used in mathematica and doesn’t have the right asymptotic properties to cancel the growth of the modified Bessel function term. However, having observed that the term should be independent of n, we can choose any n we like. So choose n=1 for the hypergeometric term. Completing the summation over the coefficient out the front we have $$f\left( x \right)\simeq -\frac{{{\pi }^{2}}}{8}\left( \log \left( \tfrac{1}{4}{{x}^{2}} \right){{I}_{0}}\left( x \right)+2{{F}^{\left( \left\{ 0 \right\},\left\{ 1,0 \right\},0 \right)}}\left( \left\{ 1 \right\},\left\{ 1,1 \right\},\tfrac{1}{4}{{x}^{2}} \right) \right)$$ Plotting this we recover an almost identical approximation as before. In the graph below, the numerical evaluation of the integral representation(black thin), vs the above asymptotic approximation (black dashed) vs the previous approximation $f\left( x \right)\simeq {{\pi }^{5/2}}{{e}^{-x}}/\sqrt{32x}$(red thin) is plotted, this time, over a domain of $x\in \left[ 0,10 \right]$ so as to observe the divergence as well as display the slight discrepancy between both approximations.