Looking for a closed form of $I(n,m)=\int_0^{+\infty} e^{-ax^n-\frac{b}{x^m}} \, dx$ I am looking for a closed form of this integral for reals $a,b>0$ and integers $n,m>0$
$I(n,m)=\int_0^{+\infty} e^{-ax^n-\frac{b}{x^m}}$
I read about $I(2,2)$ in the book Irresistible integrals.
However, the neat proof included in the book does not lend itself to generality, or even other particular cases that I know of...
The proof for $I(2,2)$ is essentially subsituting $u=\sqrt{a}x$,  then $t=\frac{\sqrt{ab}}{u}$ to get :
$I(2,2)=\int_0^{+\infty}e^{-t^2-\frac{ab}{t^2}}\times\frac{1}{\sqrt{a}}\times\frac{1}{\sqrt{ab}}\times\frac{ab}{t^2}dt=\sqrt{b}\times\int_0^{+\infty}e^{-t^2-\frac{ab}{t^2}}\times \frac{1}{t^2}dt$
So adding this new form with the previous one (after the first subsitution) yields :
$2I(2,2)=\frac{1}{\sqrt{a}}\int_0^{+\infty}e^{-t^2-\frac{ab}{t^2}}\times(\frac{\sqrt{ab}}{t^2}+1)dt=\frac{1}{\sqrt{a}}\int_0^{+\infty}e^{-t^2-\frac{ab}{t^2}}\times d(t-\frac{\sqrt{ab}}{t})$
After completing the square in the exponential, and using $\int_{-\infty}^{+\infty}e^{-x^2}=\sqrt{\pi}$, one eventually gets :
$I(2,2)=\frac{1}{2} \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}$
Trying the same method and expecting to be able to "complete the cube" for the $3,3$ case is already not working for me (I can't get the integral to simplify to something I know already)
I think the probability for a closed form to exist for all $n,m$ is pretty low... but is there a closed form for some particular cases ? Like $n=3, m=3$ ? 
 A: An integral representation of the modified Bessel function of the second kind  is $$K_{v}(z) = \int_{0}^{\infty} e^{-z \cosh t} \cosh(vt) \, dt = \frac{1}{2}\int_{-\infty}^{\infty} e^{-z \cosh t} e^{-vt} \, dt \, ,  \quad \text{Re}(z) >0. $$
A derivation can be found in chapter 6, section 22 of the classical textbook A Treatise on the Theory of Bessel Functions  by G.N. Watson.
At the end of the section, Watson mentions that by making the substitution $ \tau = \frac{z}{2} e^{t}$, we get the alternative representation
$$K_{v}(z) = \frac{1}{2} \left(\frac{z}{2} \right)^{v} \int_{0}^{\infty} \frac{e^{-\tau-z^{2}/(4 \tau)}}{\tau^{v+1}} \, d \tau \, , \quad \text{Re}(z^{2}) >0. $$
Therefore, when $m=n$, we have $$ \begin{align}I(n,n) &=  \int_{0}^{\infty} e^{-at^{n}-b/t^{n}} \, dt \\ &= \frac{1}{n}\int_{0}^{\infty} e^{-au-b/u}  u^{1/n-1} \, du \tag{1}\\ &= \frac{1}{n} \int_{0}^{\infty} e^{-w-ab/w} \left(\frac{w}{a}\right)^{1/n-1} \frac{dw}{a} \tag{2} \\ &= \frac{1}{na^{1/n}} \int_{0}^{\infty} \frac{e^{-w-(2\sqrt{ab})^{2}/(4w)}}{w^{-1/n+1}} \, dw \\ &= \frac{1}{na^{1/n}} \,  2 \left(\frac{2 \sqrt{ab}}{2} \right)^{1/n}K_{-1/n} \left(2 \sqrt{ab} \right) \\ &= \frac{2}{n}  \left(\frac{b}{a} \right)^{1/(2n)} K_{1/n} (2 \sqrt{ab}) \tag{3}. \end{align}$$
Unless $n =2$, $(3)$ cannot be expressed in terms of elementary functions.
But if $n=3$, it can be expressed in terms of the Airy function.
Specifically, 
$$I(3,3) = \frac{2}{3}  \left(\frac{b}{a} \right)^{1/6} K_{1/3} (2 \sqrt{ab}) = \frac{2 \pi}{3^{5/6}\sqrt[3]{a}} \, \text{Ai} \left(3^{2/3} \sqrt[3]{ab} \right).$$

$(1)$ Let $u=t^{n}$.
$(2)$ Let $w = au$.
$(3)$ $K_{-v}(z) = K_{v}(z)$
