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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!

robjohn
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Timtam
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    Usually in the saddle point method, there is a large parameter in the exponential in the integrand. Is it $n$? – Ron Gordon Jul 11 '13 at 01:47
  • yes I edited to put that in – Timtam Jul 11 '13 at 01:49
  • I think that Ron Gordon meant a large parameter in the exponent of $e^{-ax^2-bx^4/2}$ – robjohn Jul 11 '13 at 03:00
  • Apparently, I am just missing the point. Since the integrand is real, we can just integrate along the real axis. There is no reason to even get involved with contour integration here... I'm still not 100% sure I understand this though... if anyone wants to take a shot at it – Timtam Jul 11 '13 at 01:47
  • By the change $t=x^2$ the integral under consideration is reduced to 3.462.1 from Gradshtein and Ryzhik. Mathematica 9 also calculates it as$$ 2^{-n/2-5/4} b^{-n/2-1/4} \Gamma(n+1/2)HypergeometricU(n/2+1/4,1/2,a^2/(2 b)).$$ – user64494 Jul 11 '13 at 08:13
  • @robjohn: sorry, no, I meant large $n$, as I think the OP did as well. – Ron Gordon Jul 11 '13 at 18:37

3 Answers3

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$$ \int_0^\infty x^{2n}e^{-ax^2-bx^4/2}\,\mathrm{d}x $$ 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 $$ \sqrt{\frac{\pi\vphantom{A}}{2P''(u_0)}}\,e^{-P(u_0)} $$

robjohn
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    Can you give a reference? Usually the Laplace method is applied to the integrals of the form $$\int_a^b e^{Mf(x)},dx,$$ where $M$ is a large number. – user64494 Jul 12 '13 at 03:42
  • @user64494: This is not of that form. The location of the maximum of the exponent get larger with $n$. We just write $$P(u)\doteq P(u_0)+\frac12P''(u_0)(u-u_0)^2$$ and we get the integral given. – robjohn Jul 12 '13 at 05:25
  • @ robjohn : Unfortunaly, your calculations are not based. N. G. de Bruijn considers a modified Gamma function on a dozen of pages. M. Fedoryuk's proof of the asymptotics depending on a parameter also takes a few pages. – user64494 Jul 12 '13 at 06:36
  • @user64494: the first term approximation appeared to be what the OP was looking for. If we want to get more terms in the asymptotic expansion, I agree that more care needs to be taken. My approximation matches the saddle point approximation given at the bottom of page 17 of Section 5. – robjohn Jul 12 '13 at 08:03
  • @ robjohn : Does the asymptotic expansion given by you coincide with (2) at the bottom of page 17 of the indefinite reference? – user64494 Jul 12 '13 at 08:25
  • @user64494: Looking at $(2)$ on page 17, I see that they ignore the contribution of $g$ to the saddle point. In the case at hand, $(2)$ differs from my asymptotic expansion by a factor of $\sqrt2$. However, the method I use is the same: approximate $P(u)=P(u_0)+P''(u_0)(u-u_0)^2/2$ around $u_0$, where $P'(u_0)=0$. I can trace the factor to the fact that $P''(u_0)\to2b$ whereas $(2)$ computes it to be $b$ by ignoring the congtribution of $g$. – robjohn Jul 12 '13 at 17:54
  • Suggestion to the answer (v1): To get the $u$-integral on the form of Laplace's method (up to an irrelevant subdominant $a$-term), make a substitution $u=\sqrt{m}v$. (This substitution does not change the result; it is just for justification of the approximative formula.) – Qmechanic Oct 21 '16 at 10:13
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$\int_0^\infty x^{2n}e^{-ax^2-\frac{b}{2}x^4}~dx$

$=\int_0^\infty x^{2n}e^{-x^2\left(a+\frac{b}{2}x^2\right)}~dx$

$=\int_0^\infty\left(\dfrac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^{2n}e^{-\left(\frac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^2\left(a+\frac{b}{2}\left(\frac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^2\right)}~d\left(\dfrac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)$

$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{2a\sinh^2x(a+a\sinh^2x)}{b}}\sinh^{2n}x\cosh x~dx$

$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{2a^2\sinh^2x\cosh^2x}{b}}\left(\dfrac{e^x-e^{-x}}{2}\right)^{2n}\dfrac{e^x+e^{-x}}{2}dx$

$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\sinh^22x}{2b}}\dfrac{e^x+e^{-x}}{2}\left(\dfrac{(-1)^nC_n^{2n}}{4^n}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}C_{n-k}^{2n}(e^{2kx}+e^{-2kx})}{4^n}\right)dx$

$=\dfrac{\sqrt2a^{n+\frac{1}{2}}}{2^nb^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2}{2b}\frac{\cosh4x-1}{2}}\left(\dfrac{(-1)^n(2n)!(e^x+e^{-x})}{2(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(e^{2kx}+e^{-2kx})(e^x+e^{-x})}{2(n+k)!(n-k)!}\right)dx$

$=\dfrac{\sqrt2a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^nb^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh4x}{4b}}\left(\dfrac{(-1)^n(2n)!(e^x+e^{-x})}{2(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(e^{(2k+1)x}+e^{-(2k-1)x}+e^{(2k-1)x}+e^{-(2k+1)x})}{2(n+k)!(n-k)!}\right)dx$

$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh4x}{4b}}\left(\dfrac{(-1)^n(2n)!\cosh x}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(\cosh((2k+1)x)+\cosh((2k-1)x))}{(n+k)!(n-k)!}\right)d(4x)$

$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh x}{4b}}\left(\dfrac{(-1)^n(2n)!\cosh\dfrac{x}{4}}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!\left(\cosh\dfrac{(2k+1)x}{4}+\cosh\dfrac{(2k-1)x}{4}\right)}{(n+k)!(n-k)!}\right)dx$

$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\left(\dfrac{(-1)^n(2n)!K_\frac{1}{4}\left(\dfrac{a^2}{4b}\right)}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!\left(K_\frac{2k+1}{4}\left(\dfrac{a^2}{4b}\right)+K_\frac{2k-1}{4}\left(\dfrac{a^2}{4b}\right)\right)}{(n+k)!(n-k)!}\right)$

Harry Peter
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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).

user64494
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