1
$\begingroup$

How can I calculate the integral $$\int_0^\infty{\frac1y e^{\frac{-x_0}y-y}}dy$$ in terms of well-known constants and functions?

I used some fundamental techniques of integration but got nothing.

$\endgroup$
4
  • $\begingroup$ This is an integration with respect to $y$? $\endgroup$ – Spencer Jan 5 '15 at 0:35
  • $\begingroup$ yes that's true... $\endgroup$ – k1.M Jan 5 '15 at 0:36
  • $\begingroup$ Did you succeed??? $\endgroup$ – k1.M Jan 5 '15 at 0:50
  • $\begingroup$ I used some fundamental techniques but got nothing - Not surprisingly, since the integral can only be expressed in terms of the special Bessel function. In particular, $2K_0(2\sqrt x)$. $\endgroup$ – Lucian Jan 5 '15 at 0:54
2
$\begingroup$

Let $u = y+x_0/y$, then

$$y=\frac12 \left (u \pm \sqrt{u^2-4 x_0} \right ) $$ $$dy=\frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 x_0}} \right ) du $$

Then the integral may be rewritten as (see this answer)

$$ \int_{\infty}^{2 \sqrt{x_0}} du \left (1 - \frac{u}{\sqrt{u^2-4 x_0}} \right ) \frac{e^{-u}}{\left (u - \sqrt{u^2-4 x_0} \right )} + \int_{2 \sqrt{x_0}}^{\infty} du \left (1 + \frac{u}{\sqrt{u^2-4 x_0}} \right ) \frac{e^{-u}}{\left (u + \sqrt{u^2-4 x_0} \right )} $$

which simplifies to

$$\begin{align}2 \int_{2 \sqrt{x_0}}^{\infty} du \, \left (u^2-4 x_0 \right )^{-1/2} e^{-u} &= 2 \int_1^{\infty} dv \, (v^2-1)^{-1/2} e^{-2 \sqrt{x_0} v}\\ &= 2 \int_0^{\infty} dt \, e^{-2 \sqrt{x_0} \cosh{t}} \\ &= 2 K_0 \left ( 2 \sqrt{x_0}\right ) \end{align}$$

where $K_0$ is the modified Bessel function of the second kind of zeroth order.

$\endgroup$
1
  • $\begingroup$ thanx! nice solution... $\endgroup$ – k1.M Jan 5 '15 at 0:55
2
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{1 \over y}\,\exp\pars{-\,{x_{0} \over y} - y}\,\dd y}\ =\ \overbrace{\int_{0}^{\infty}{1 \over y}\,\exp\pars{-\root{x_{0}}\, \bracks{{\root{x_{0}} \over y} + {y \over \root{x_{0}}}}}\,\dd y} ^{\ds{\dsc{y}=\dsc{\root{x_{0}}\exp\pars{\theta}}}} \\[5mm]&=\int_{-\infty}^{\infty}{1 \over \root{x_{0}}\expo{\theta}}\, \exp\pars{-\root{x_{0}}\bracks{\expo{-\theta} + \expo{\theta}}}\,\root{x_{0}}\expo{\theta}\,\dd\theta \\[5mm]&=2\int_{0}^{\infty}\exp\pars{-2\root{x_{0}}\cosh\pars{\theta}}\,\dd\theta =\color{#66f}{\large 2\,{\rm K}_{0}\pars{2\root{x_{0}}}} \end{align}

$\ds{\,{\rm K}_{\nu}}$ is a Bessel Function. See $\ds{\bf 9.6.24}$ in this link.

$\endgroup$

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.