Saddle Point Integral I want to calculate , 
$$I = \int_0^\infty dx \,x^{2n}e^{-ax^2 -\frac{b}{2}x^4} $$
for real positive a, b and positive integer n. n is the large parameter. Using Saddle Point Integration
I find saddle points  by setting the derivative  P'(x) = 0 where
$$ P(x) = n\log(x^2) -ax^2 -\frac{b}{2}x^4$$
In order to do this I never know which saddle point to use !  I see there are two imaginary ones and two real ones. I think I want the one that is positive and real but I have no idea  why. (My professor hinted at this one).
By the way the reason there are two real solutions and two imaginary ones is actually not completely obvious to me but I believe that is the case by inspecting the function you get
$$ 0 =  n -ax^2 -bx^4$$
this function has two real roots so the other two must be imaginary.
My professor said: "Just plot the integrand at positive
psi and you will see what saddle point to use"
I looked at the plot using coefficient n=a=b= 1 but i didn't get how that tells me which saddle point to use.
Any help would be appreciated !thanks!
 A: $$
\int_0^\infty x^{2n}e^{-ax^2-bx^4/2}\,\mathrm{d}x
$$
Let $x=x^2$ and $m=n-1/2$, and we get
$$
\frac12\int_0^\infty u^me^{-au-bu^2/2}\,\mathrm{d}u
=\frac12\int_0^\infty e^{-P(u)}\,\mathrm{d}u
$$
where
$$
\begin{align}
P(u)&=bu^2/2+au-m\log(u)\\
P'(u)&=bu+a-m/u\\
P''(u)&=b+m/u^2
\end{align}
$$
We get $P'(u_0)=0$ for
$$
u_0=\frac{-a+\sqrt{a^2+4bm}}{2b}
$$
and
$$
P''(u_0)=\frac{a^2+4bm+a\sqrt{a^2+4bm}}{2m}
$$
The Saddle Point method gives the asymptotic approximation
$$
\sqrt{\frac{\pi\vphantom{A}}{2P''(u_0)}}\,e^{-P(u_0)}
$$
A: $\int_0^\infty x^{2n}e^{-ax^2-\frac{b}{2}x^4}~dx$
$=\int_0^\infty x^{2n}e^{-x^2\left(a+\frac{b}{2}x^2\right)}~dx$
$=\int_0^\infty\left(\dfrac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^{2n}e^{-\left(\frac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^2\left(a+\frac{b}{2}\left(\frac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^2\right)}~d\left(\dfrac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)$
$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{2a\sinh^2x(a+a\sinh^2x)}{b}}\sinh^{2n}x\cosh x~dx$
$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{2a^2\sinh^2x\cosh^2x}{b}}\left(\dfrac{e^x-e^{-x}}{2}\right)^{2n}\dfrac{e^x+e^{-x}}{2}dx$
$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\sinh^22x}{2b}}\dfrac{e^x+e^{-x}}{2}\left(\dfrac{(-1)^nC_n^{2n}}{4^n}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}C_{n-k}^{2n}(e^{2kx}+e^{-2kx})}{4^n}\right)dx$
$=\dfrac{\sqrt2a^{n+\frac{1}{2}}}{2^nb^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2}{2b}\frac{\cosh4x-1}{2}}\left(\dfrac{(-1)^n(2n)!(e^x+e^{-x})}{2(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(e^{2kx}+e^{-2kx})(e^x+e^{-x})}{2(n+k)!(n-k)!}\right)dx$
$=\dfrac{\sqrt2a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^nb^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh4x}{4b}}\left(\dfrac{(-1)^n(2n)!(e^x+e^{-x})}{2(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(e^{(2k+1)x}+e^{-(2k-1)x}+e^{(2k-1)x}+e^{-(2k+1)x})}{2(n+k)!(n-k)!}\right)dx$
$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh4x}{4b}}\left(\dfrac{(-1)^n(2n)!\cosh x}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(\cosh((2k+1)x)+\cosh((2k-1)x))}{(n+k)!(n-k)!}\right)d(4x)$
$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh x}{4b}}\left(\dfrac{(-1)^n(2n)!\cosh\dfrac{x}{4}}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!\left(\cosh\dfrac{(2k+1)x}{4}+\cosh\dfrac{(2k-1)x}{4}\right)}{(n+k)!(n-k)!}\right)dx$
$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\left(\dfrac{(-1)^n(2n)!K_\frac{1}{4}\left(\dfrac{a^2}{4b}\right)}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!\left(K_\frac{2k+1}{4}\left(\dfrac{a^2}{4b}\right)+K_\frac{2k-1}{4}\left(\dfrac{a^2}{4b}\right)\right)}{(n+k)!(n-k)!}\right)$
A: The required mathematical background can be seen in N. G de Bruijn, Asymptotic methods in analysis, North-Holland Publ. Co - Amsterdam, P. Noordhoff LTD - Groningen (1958), Ch. 6, 6.8 A modified Gamma function. The asymptotics of the integral to which the integral under consideration can be easily reduced  (together with the proofs) is contained in M. Fedoryuk, Saddle method, Nauka, Moscow (1977) (in Russian). 
