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.
$\begingroup$
$\endgroup$
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
Add a comment
|
$\begingroup$
$\endgroup$
1
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.
answered Jan 5 '15 at 0:51
$\begingroup$
$\endgroup$
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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.
answered Jan 7 '15 at 10:14
