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