4
$\begingroup$

I'm practicing harder integration using techniques of solving with special functions

I have difficulties with these two hard integrals; don't even know how to start,

$$\int_0 ^\infty x^p e^{-\frac{\theta}{x}+Bx}dx$$ where $\theta,B>0$

$$pv\int_0 ^\infty \frac{x^p e^{\cos{\theta x}} \cos(\frac{\pi p}{2} - \sin\theta x)}{1-x^2}dx$$

please help me to start of giving your solutions! thank you so much and have a good day/night

$\endgroup$
  • $\begingroup$ Ohh... brilliant user .. ..... $\endgroup$ – Aman Rajput Jun 29 '16 at 16:44
  • $\begingroup$ You should post them as a note there also.. hope someone will help you $\endgroup$ – Aman Rajput Jun 29 '16 at 16:45
  • 2
    $\begingroup$ What about this beautiful thing called $\LaTeX$? $\endgroup$ – PHPirate Jun 29 '16 at 16:54
  • 1
    $\begingroup$ Might be better to cut this into two questions. $\endgroup$ – mvw Jun 29 '16 at 17:01
  • 1
    $\begingroup$ Then the first integral doesn't converge. You have an $\exp(+Bx)$ factor that grows ever larger as $x\rightarrow\infty$. $\endgroup$ – John Barber Jun 29 '16 at 17:42
5
$\begingroup$

Let $\theta=b$ and $B = -a \lt 0$. Then the first integral is a generalization of this integral. Using the same substitutions:

$u=a x+b/x$. Then $a x^2-u x+b = 0$, and therefore

$$x = \frac{u}{2 a} \pm \frac{\sqrt{u^2-4 a b}}{2 a}$$

$$dx = \frac1{2 a}\left (1 \pm \frac{u}{\sqrt{u^2-4 a b}}\right ) du $$

Now, it should be understood that as $x$ traverses from $0$ to $\infty$, $u$ traverses from $\infty$ down to a min of $2 \sqrt{a b}$ (corresponding to $x \in [0,\sqrt{b/a}]$), then from $2 \sqrt{a b}$ back to $\infty$ (corresponding to $x \in [\sqrt{b/a},\infty)$). Therefore the integral is

$$\frac1{2 a} \int_{\infty}^{2 \sqrt{a b}} du \, \left (\frac{u}{2 a} - \frac{\sqrt{u^2-4 a b}}{2 a} \right )^p \left (1 - \frac{u}{\sqrt{u^2-4 a b}}\right ) e^{-u} \\+ \frac1{2 a} \int_{2 \sqrt{a b}}^{\infty} du \, \left (\frac{u}{2 a} + \frac{\sqrt{u^2-4 a b}}{2 a} \right )^p \left (1 + \frac{u}{\sqrt{u^2-4 a b}}\right ) e^{-u} $$

which simplifies, subbing $u=2 \sqrt{a b} \cosh{v}$, to

$$\begin{align}\int_0^{\infty} dx \, x^p e^{-\left (a x+\frac{b}{x} \right )} &=2 \left (\sqrt{\frac{b}{a}} \right )^{p+1} \int_0^{\infty} dv \, \cosh{[(p+1) v]} \, e^{-2 \sqrt{a b} \cosh{v}} \\ &=2 \left (\sqrt{\frac{b}{a}} \right )^{p+1} K_{p+1} \left ( 2 \sqrt{a b}\right )\end{align}$$

$\endgroup$
4
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[2]{\,\mathrm{Li}_{#1}\left(\,{#2}\,\right)} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\theta > 0}$ and $\ds{B > 0}$, $\ds{\int_{0}^{\infty}x^{p} \exp\pars{-\,{\theta \over x} \color{#f00}{\large -} Bx}\,\dd x =\, ?}$.


Write $\ds{x \equiv \root{\theta \over B}\expo{t}}$ such that

\begin{align} &\color{#f00}{\int_{0}^{\infty}x^{p} \exp\pars{-\,{\theta \over x} - Bx}\,\dd x} = \int_{-\infty}^{\infty}\pars{\root{\theta \over B}\expo{t}}^{p} \exp\pars{-\root{\theta B}\bracks{\expo{-t} + \expo{t}}}\root{\theta \over B} \expo{t}\,\dd t \\[3mm] = &\ \pars{\theta \over B}^{\pars{p + 1}/2}\int_{-\infty}^{\infty} \expo{\pars{p + 1}t}\exp\pars{-2\root{\theta B}\cosh\pars{t}}\,\dd t \\[3mm] = &\ \pars{\theta \over B}^{\pars{p + 1}/2}\int_{-\infty}^{\infty} \braces{\vphantom{\Large A}\cosh\pars{\vphantom{\large A}\bracks{p + 1}t} + \sinh\pars{\vphantom{\large A}\bracks{p + 1}t}} \exp\pars{-2\root{\theta B}\cosh\pars{t}}\,\dd t \\[3mm] = &\ 2\pars{\theta \over B}^{\pars{p + 1}/2}\int_{0}^{\infty} \cosh\pars{\bracks{p + 1}t}\exp\pars{-2\root{\theta B}\cosh\pars{t}}\,\dd t \\[3mm] = &\ \color{#f00}{2\pars{\theta \over B}^{\pars{p + 1}/2}\, \mathrm{K}_{p + 1}\pars{2\root{\theta B}}} \end{align}


$\ds{\mathrm{K}_{\nu}\pars{z}}$ is the Modified Bessel Function of the Second Kind.

$\endgroup$

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.