Asymptotic power series expansion of $\int_0^\infty\frac{x^\nu J_\nu(x\alpha)}{e^x-1}{\rm d}x$ around $\alpha=1$ and $\alpha<1$

What is the power series expansion $$f(\alpha)$$ as $$\alpha\to 1$$ and also in the limit $$\alpha\ll 1$$, where

$$f(\alpha)=\int_0^\infty\frac{x^\nu J_\nu(x\alpha)}{e^x-1}{\rm d}x$$

At $$\alpha=1$$ the integrand is analytic so we should have a convergent power series. The power series expansion for the case $$\alpha\gg1$$ is already answered here. How does this differ from the expansion at $$\alpha=1$$ and in the region $$\alpha<1$$.

I see that the numerical plot ($$\nu=2$$) of the integration is the following

Thus something very interesting is happening at the point $$\alpha=1$$.

Edit: If it is difficult to find an asymptotic expansion at $$\alpha=1$$, then at least can we prove that $$f(\alpha)$$ has a maxima at that point?

• Letting $$F_{\nu}(s)=\int_0^\infty\frac{x^\nu J_{\nu}(sx)}{e^x-1}\mathrm dx$$ We can see by the Leibniz rule that $${F_{\nu}}'(1)=\int_0^\infty \frac{x^{\nu+1}{J_{\nu}}'(x)}{e^x-1}\mathrm dx$$ So show this is $0$ $\forall \nu$ (easier said than done). Trying to use the recurrences for the Bessel function might help. Oct 23, 2021 at 13:04
• Set $x=2\pi t$ and use the Abel Plana formula. Oct 25, 2021 at 1:31

For $$a$$ near 1 the 'multiplication theorem' (see wiki on Bessel functions) can be used:

$$J_\nu(a \ x) = a^\nu \sum_{n=0}^\infty \frac{1}{n!} \Big( \frac{(1-a^2)x}{2}\Big)^n J_{\nu+n}(x)$$

Keeping the $$n=0$$ and $$n=1$$ terms gives $$\int_0^\infty \frac{x^\nu J_\nu(a \ x)}{e^x-1} dx\sim a^\nu \big(C_0^\nu+C_1^\nu(1-a^2)/2 \big)$$ $$C_0^\nu= \int_0^\infty \frac{x^\nu J_\nu( x)}{e^x-1}dx \ ,\quad C_1^\nu= \int_0^\infty \frac{x^{\nu+1} J_{\nu+1}( x)}{e^x-1}dx$$ This gives a very good approximation near $$a=1.$$ For instance, with $$a=0.99,$$ and $$\nu = 0.4,$$ the two-term approximation is 0.0037% different from the actual and for $$\nu = 0.8,$$ 0.0096% different. Of course, if you have to calculate numerically two integrals at exactly $$a=1,$$ there seems to be little practical advantage. The exception to this is if you are doing some theoretical work and you need the analytic behavior as $$a \to 1,$$ and you really don't care so much for the exact values of $$C_0^\nu$$ and $$C_1^\nu,$$ but that they exist.

• Yes, I'm doing a theoretical work for which only the asymptotic form is important. Can you please confirm whether we can ignore the $C^\nu_1$ term while only keeping the $C^\nu_0$ term? Oct 26, 2021 at 16:56
• @FaberBosch For the cases of $\nu=0.4$ and $\nu=0.8,$ both have an error less than 1%. It seems the larger $\nu$ is, the worse the approximation. (I've assumed fixed $\nu$ in the analysis.) I'd think you'd want the first correction term, because the $a^v$ dependence is weak for $a \to 1.$ It's that $(1-a^2)/2 \sim (1-a)$ that shows successive terms will be small. Oct 26, 2021 at 17:23
• Note that $$C_n^\nu = \int_0^{ + \infty } {\frac{{x^{\nu + n} }}{{e^x - 1}}J_{\nu + n} (x)dx = } \frac{{2^{\nu + n} }}{{\sqrt \pi }}\Gamma \!\left( {\nu + n + \tfrac{1}{2}} \right)\sum\limits_{k = 1}^\infty {\frac{1}{{(k^2 + 1)^{\nu + n + 1/2} }}} .$$
– Gary
Oct 28, 2021 at 4:39
• This formula for $C_n^\nu$ shows that the $n$th term in skbmoore's series behaves like $$\frac{1}{{\sqrt {2\pi } }}n^{\nu - \frac{1}{2}} \left( {\frac{{1 - a^2 }}{2}} \right)^n .$$ Thus, the series converges inside the Cassini oval $\left| {1 - a^2 } \right|^2 = 4$. The presence of $a^\nu$ may require the removal of the segment $[ - \sqrt 3 ,0]$ from this oval.
– Gary
Oct 28, 2021 at 6:51
• @Gary Nice observations. Oct 28, 2021 at 16:28

Via expanding the series for the Bessel function and then integrating term-wise, one gets: $$\frac{(2a)^v}{\sqrt{\pi}} \sum_{m=0}^{\infty} (-1)^m \frac{\Gamma(m+v+1/2)}{\Gamma(m+1)} \zeta(2m+2v+1) \: a^{2m}$$

where $$\zeta(m)$$ is the Riemann zeta function.

This, however, is only convergent for $$a < 1$$ though.

Let us use the Abel-Plana formula:

$$\sum_0^\infty f(x)=\frac12 f(0)+\int_0^\infty f(x)dx+i\int_0^\infty \frac{f(-ix)-f(ix)}{e^{2\pi x}-1 }dx$$

which works for “weak bounds” as stated in the article. Let $$f(x)=\frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa)$$

of which the Modified Bessel Functions of the First Kind simplify as $$f(-ix)-f(ix)=(2\pi)^v x^v \text J_v(2\pi xa)$$.

Therefore:

$$\sum_{x=0}^\infty \frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) =\frac12 \frac i2\csc(\pi v)(2\pi 0)^v\text I_v(2\pi 0a) +\int_0^\infty \frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) dx+i\int_0^\infty \frac{\frac i2\csc(\pi v)(-2\pi ix)^v\text I_v(-2\pi i xa) -\frac i2\csc(\pi v)(2\pi ix)^v\text I_v(2\pi ixa)}{e^{2\pi x}-1 }dx\implies i\int_0^\infty \frac{(2\pi)^v x^v \text J_v(2\pi xa)}{e^{2\pi x}-1 }dx+ \int_0^\infty \frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) dx = \sum_{x=0}^\infty\frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa)$$ Rearranging:

$$i\int_0^\infty \frac{(2\pi)^v x^v \text J_v(2\pi xa)}{e^{2\pi x}-1 }dx+ \int_0^\infty \frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) dx = \sum_{x=0}^\infty\frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) \implies i\int_0^\infty \frac{(2\pi)^v x^v \text J_v(2\pi xa)}{e^{2\pi x}-1 }dx = \sum_{x=0}^\infty\frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) -\int_0^\infty \frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) dx$$

Factoring and substituting $$2\pi x=t\implies \frac{dt}{2\pi}= dx,t_1=2\pi 0=0,t_2=2\pi\infty=\infty$$:

$$i\int_0^\infty \frac{(2\pi)^v x^v \text J_v(2\pi xa)}{e^{2\pi x}-1 }dx = \sum_{x=0}^\infty\frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) -\int_0^\infty \frac i2\csc(\pi v)(2\pi x)^v\text I_v(2\pi xa) dx\mathop\implies ^{2\pi x=t}\frac i{2\pi}\int_0^\infty \frac{t^v \text J_v(ta)}{e^t-1 }dt = \sum_{\frac t{2\pi}=0}^\infty\frac i2\csc(\pi v)t^v\text I_v(ta) -\frac1{2\pi}\int_0^\infty \frac i2\csc(\pi v)t^v\text I_v(ta) dt$$

Factoring and some algebra:

$$\int_0^\infty \frac{t^v \text J_v(ta)}{e^t-1 }dt =- 2\pi i\sum_{\frac t{2\pi}=0}^\infty\frac i2\csc(\pi v)t^v\text I_v(ta) - -2\pi i\frac1{2\pi}\int_0^\infty \frac i2\csc(\pi v)t^v\text I_v(ta) dt= 2\pi i \frac i2\csc(\pi v)\sum_{\frac t{2\pi}=0}^\infty t^v\text I_v(ta) - -2\pi i\frac1{2\pi} \frac i2\csc(\pi v)\int_0^\infty t^v\text I_v(ta) dt =-\csc(\pi v)\left[\pi\sum_{\frac t{2\pi}=0}^\infty t^v\text I_v(ta)+\frac12 \int_0^\infty t^v\text I_v(ta) dt\right]$$

So our final result according to the Abel-Plana formula is: $$f(a)= \sum_0^\infty f(x)=\frac12 f(0)+\int_0^\infty f(x)dx+i\int_0^\infty \frac{f(-ix)-f(ix)}{e^{2\pi x}-1 }dx=-\csc(\pi v)\left[\pi\sum_{\frac x{2\pi}=0}^\infty x^v\text I_v(xa)+\frac12 \int_0^\infty x^v\text I_v(xa) dx\right]= -\csc(\pi v)\left[2^v\pi^{v+1}\sum_{ x=0}^\infty x^v\text I_v(2\pi ax)+\frac12 \int_0^\infty x^v\text I_v(xa) dx\right]$$

Note that $$\int x^v\text I_v(ax)dx=a^v \frac{x^{2v+1}}{2^{v+1}}\Gamma\left(v+\frac12\right)\,_1\tilde{\text F}_2\left(v+\frac12;v+1,v+\frac32;\frac{a^2 x^2}4\right)+C$$

The result may be a difference of large numbers.I hope this helps. You can also try this special case geometric series expansion. Please correct me and give me feedback!

• The series and integrals will not converge because $I_\nu$ grows exponentially. Note that the usual notation for the order of Bessel functions is nu: $\nu$ and not a v: $v$.
– Gary
Oct 25, 2021 at 3:41
• Any comments on my observation?
– Gary
Oct 28, 2021 at 4:39
• @Gary Would it be possible to use the Abel-Plana formula in another way or do it’s conditions not let it? Oct 28, 2021 at 12:18
• You may not be able to solve this problem using A-P. I observed from your recent posts that you are quite obsessed with A-P, and try to approach all summation problems with it. It is a nice formula indeed but not a universal answer to all your problems. dlmf.nist.gov/2.10.E2 tells you the conditions for it to hold, please check these whenever you are trying to use A-P. If the conditions are clearly not met, please do not post a result using A-P.
– Gary
Oct 28, 2021 at 23:56
• No problem with that. I am just saying that in many cases, unfortunately, it just does not work.
– Gary
Oct 29, 2021 at 0:08