3
$\begingroup$

What tools, ways would you propose for getting the closed form of this integral?

$$\int^\infty_{B}e^{-\left(x+\frac{A}{x}\right)}\,dx,$$ where $A>0$, $B>0$.

When $B=0$, from Table of Integrals, Series, and Products, we can use this to calculate $$\int^\infty_0 \exp \left(-\frac{\beta}{4x}-\gamma x\right)\,dx= \sqrt{\frac{\beta}{\gamma}} K_1(\sqrt{\beta\gamma})$$ with $\operatorname{Re} \beta\geq 0$, $\operatorname{Re} \gamma > 0 $.

But when $B>0$, how to calculate this intergral?

Thank you everyone.

$\endgroup$
  • $\begingroup$ These look at least as complicated as Bessel functions (take $B=0$), so it is unlikely they have a "nice" closed form. I recommend numerical tools for computation. $\endgroup$ – parsiad Feb 5 '16 at 15:53
  • $\begingroup$ Can this intergral be expressed by using Bessel functions ? @par $\endgroup$ – Leung Feb 5 '16 at 16:08
  • $\begingroup$ Another data point.... $$\int_1^\infty \exp(-(x+\frac{1}{x}))\;dx \approx 0.20753352343482877323 $$ is not recognized by the ISC. $\endgroup$ – GEdgar Feb 5 '16 at 16:13
  • $\begingroup$ what is the 'ISC' ? @GEdgar. I have calculated some data points by using Matlab. But I want to find the closed form of this integral. $\endgroup$ – Leung Feb 5 '16 at 16:24
  • $\begingroup$ @mk4201 ... Inverse Symbolic Calculator isc.carma.newcastle.edu.au For example: Compute this integral numerically for $B=0$, plug that number into ISC, and get the Bessel answer. $\endgroup$ – GEdgar Feb 5 '16 at 16:26
3
$\begingroup$

Let $u=x+A/x$ so that $x^2-u x+A=0$, or

$$x = \frac{u}{2} \pm \frac12 \sqrt{u^2-4 A} $$ $$dx = \frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 A}} \right ) du$$

Note that the integration limits provided by the mapping $x \mapsto u$ depend on whether the point $x=B$ is less than or greater than the minimum of $u$ at $x=\sqrt{A}$. Let's assume the former, i.e., $B \lt \sqrt{A}$; then the integral over $u$ is

$$\begin{align} I(A,B) &= \frac12 \int_{B+A/B}^{2 \sqrt{A}} du \, \left (1 - \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} + \frac12 \int_{2 \sqrt{A}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u}\\ &= \frac12 e^{-\left (B+\frac{A}{B} \right )} + 2 \sqrt{A} K_1 \left (2 \sqrt{A} \right ) - \sqrt{A}\int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}} \end{align} $$

where

$$\alpha = \operatorname{arccosh}{\left (\frac{B+\frac{A}{B}}{2 \sqrt{A}} \right )}$$

If however $B \ge \sqrt{A}$,

$$\begin{align} I(A,B) &= \frac12 \int_{B+A/B}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} \\ &= \frac12 \int_{B+A/B}^{2 \sqrt{A}} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} + \frac12 \int_{2 \sqrt{A}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} \\ &= \frac12 e^{-\left (B+\frac{A}{B} \right )} + \sqrt{A} \int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}}\end{align} $$

For either case, let's consider

$$\begin{align} \int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}} &= \int_{\alpha}^{\infty} dt \, \coth{t} \sinh{t} \; e^{-2 \sqrt{A} \cosh{t}} \\ &= \frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} - \frac1{2 \sqrt{A}}\int_{\alpha}^{\infty} dt \, \frac1{\sinh^2{t}} e^{-2 \sqrt{A} \cosh{t}}\\ &= \frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} - \frac1{4 A} \frac1{\sinh^3{\alpha}} \; e^{-2 \sqrt{A} \cosh{\alpha}} + \frac{3}{4 A} \int_{\alpha}^{\infty} dt \, \frac{\cosh{t}}{\sinh^4{t}} e^{-2 \sqrt{A} \cosh{t}} \end{align} $$

We could go on and develop more terms in the series by further integration by parts, but let's take stock. Recall that

$$\cosh{\alpha} = \frac{B+\frac{A}{B}}{2 \sqrt{A}} $$ $$\sinh{\alpha} = \frac{B+\frac{A}{B}}{2 \sqrt{A}} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{1/2} $$

Then

$$\frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} = \frac1{2 \sqrt{A}} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{-1/2} e^{-\left ( B+\frac{A}{B} \right )} $$

$$\frac1{4 A} \frac1{\sinh^3{\alpha}} \; e^{-2 \sqrt{A} \cosh{\alpha}} = \frac{2 \sqrt{A}}{\left ( B+\frac{A}{B} \right )^3} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{-3/2} e^{-\left ( B+\frac{A}{B} \right )} $$

Note that additional terms will produce smaller contributions, so we can stop here for the moment and appreciate two uses of this series:

1. Perturbation for $0 \lt B \lt \sqrt{A}$ from the exact Bessel function result at $B=0$:

Using the assumption that $B \gt 0$ is a small perturbation, we expand out to $O(B^3)$, requiring the first two terms of the above expansion. Any further terms requires more integration by parts. Note that, in this region, it is the $1/B$ term that dominates. We find that

$$I(A,B) = 2 \sqrt{A} K_1 \left (2 \sqrt{A} \right ) - \left [ \frac1{2 A} B^2 + \frac1{2 A^2} B^3 +O(B^4)\right ] e^{-B-\frac{A}{B}}$$

Note that the perturbation term for small $B$ is extremely small. This has been verified numerically, although I caution that the extreme smallness of the perturbation makes it difficult to verify the exact expansion terms in $B$.

2. Global approximation of $I(A,B)$ for all $B$

We may accomplish this by considering expansions for both small $B$ and large $B$. Note that we use either one of the two cases delineated above according to whether $B$ is less than or greater than $\sqrt{A}$. I will not push this further, and I anticipate some difficulties because of the exponential term in the expansion, but I believe a two-point Pade approximant may do the trick.

$\endgroup$
  • $\begingroup$ Nice answer. But it is still very difficult for me, I need to take some time to learn this. Thank you ! $\endgroup$ – Leung Feb 6 '16 at 12:14
  • $\begingroup$ @mk4201: please feel free to ask specific questions here. $\endgroup$ – Ron Gordon Feb 6 '16 at 12:15
  • $\begingroup$ Why does the integral start at ($B+1/B$) instead of ($B+A/B$)? $\endgroup$ – Leung Feb 6 '16 at 16:21
  • $\begingroup$ @mk4201: good catch. I will fix throughout. $\endgroup$ – Ron Gordon Feb 6 '16 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.