# The closed form of $\int^\infty_{B}e^{-(x+\frac{A}{x})}\,dx$, where $A>0$, $B>0$.

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.

• 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. – parsiad Feb 5 '16 at 15:53
• Can this intergral be expressed by using Bessel functions ? @par – Leung Feb 5 '16 at 16:08
• Another data point.... $$\int_1^\infty \exp(-(x+\frac{1}{x}))\;dx \approx 0.20753352343482877323$$ is not recognized by the ISC. – GEdgar Feb 5 '16 at 16:13
• 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. – Leung Feb 5 '16 at 16:24
• @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. – GEdgar Feb 5 '16 at 16:26

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.

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