# hard integral problems to solve

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 $\theta,B>0$

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

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

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}


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.