How to find the integral $\int_{0}^{\infty}\exp(- (ax+b/x))\,dx$?
How do I find $$\large\int_{0}^{\infty}e^{-\left(ax+\frac{b}{x}\right)}dx$$ where $a$ and $b$ are positive numbers?
This is not a homework question. I will be quite happy if somebody can come up with a sort of bound, like an upper bound or a lower bound of integrand.
Solution 1:
Sub $u=a x+b/x$. 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
$$\frac1{2 a} \int_{\infty}^{2 \sqrt{a b}} du \left (1 - \frac{u}{\sqrt{u^2-4 a b}}\right ) e^{-u} + \frac1{2 a} \int_{2 \sqrt{a b}}^{\infty} du \left (1 + \frac{u}{\sqrt{u^2-4 a b}}\right ) e^{-u} $$
which simplifies to
$$\frac1{a} \int_{2 \sqrt{a b}}^{\infty} du \frac{u}{\sqrt{u^2-4 a b}} e^{-u} = 2 \sqrt{\frac{b}{a}} \int_0^{\infty} dv \cosh{v} \, e^{-2 \sqrt{a b} \cosh{v}}$$
which is then
$$\int_0^{\infty} dx \, e^{-(a x+b/x)} = 2 \sqrt{\frac{b}{a}} K_1\left ( 2 \sqrt{a b}\right )$$
Solution 2:
Of course we need $a,b>0$.
Knowing that the answer is supposed to be what Claude obtained, I can confirm that. By linear change of variables, we may assume $a=1$. Also
for convenience take $b = c^2/4$. So now let
$$F(c) = \dfrac{1}{c} \int_0^\infty e^{-x - c^2/(4 x)}\ dx$$
I claim $F(c) = K_1(c)$.
A differential equation for $y(c) = K_1(c)$ is
$$ y'' + \dfrac{1}{c} y' - \left(1 - \dfrac{1}{c^2}\right) y = 0$$
If we apply that to $F(c)$ and do some simplification, we get
$$ F'' + \dfrac{1}{c} F' - \left(1 - \dfrac{1}{c^2}\right) F =
\int_0^\infty e^{-x-c^2/(4x)} \dfrac{c^2 - 4 x^2}{4 c x^2}\ dx $$
Using the change of variables $x = c^2/(4t)$, the right side becomes
$$ - \int_0^\infty e^{-t-c^2/(4t)} \dfrac{c^2 - 4 t^2}{4 c t^2}\ dt $$
and therefore is $0$.
Now the general solution of the differential equation is $A I_1(c) + B K_1(c)$
where $A$ and $B$ are arbitrary constants. $K_1(c)$ is the solution that
goes to $0$ as $c \to +\infty$ and is
asymptotic to $1/c$ as $c \to 0+$.