integration - How to evaluate the following integral.

$$P(S)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(a^2+c^2+ \lambda ~ b^2)} \delta[S-\sqrt{4b^2+(a-c)^2}] ~da~db~bc.$$

where $\lambda$ is a constant. How to Evaluate above integral ?.

I tried!, but could reach up to mid way only.

On transforming $P(S)$ to the three dimensional the spherical polar co-ordinates using $2b=r \cos \theta, a=r\sin \theta \cos \phi, c= r \sin \theta \sin \phi$ as

$$P(S)= \int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi} e^{-r^2 (\lambda (\cos^2 \theta) /4 +\sin^2\theta ( \cos^2\phi+\sin^2\phi ) ) } \delta[S-r g(\theta, \phi)]~ r^2dr \sin \theta ~d\theta~ d\phi.$$

Crashing the delta function in above, we get a $\theta, \phi$ integral

$$P(S)= A \int_{0}^{\pi/2} \int_{0}^{\pi} e^{-S^2(\lambda (\cos^2\theta) /4 +\sin^2\theta (\cos^2 \phi+\sin^2 \phi) )/ g^2(\theta, \phi} \frac{S^2}{|g[\theta, \phi)|^3} \sin \theta ~d\theta~ d\phi,$$

Where $$g(\theta,\phi)=\sqrt{1-\sin^2\theta \sin 2 \phi}.$$ and $A$ is evaluated value (I am taking it as a constant) of $r$ integral.

After that, I took help of mathematica to evaluate it numerically.

It would be really a great help If anyone can help me by evaluating integral

Answer

First, let's deal with the delta function. We will use the delta function identity
$$\delta[f(x)]=\sum_n\frac{\delta(x-x_n)}{|f'(x_n)|},$$
where $x_n$ is a zero of the function $f(x)$ and the sum runs over all zeros. In our case,
$$f(b)=S-\sqrt{4b^2+(a-c)^2},$$
which has two zeros
$$b_{\pm}=\pm\frac{1}{2}\sqrt{S^2-(a-c)^2},$$
and the derivative of $f(b)$ can be worked out easily. Applying the above identity gives
$$\delta\left[S-\sqrt{4b^2+(a-c)^2}\right]=\frac{S}{4|b|}\left(\delta\left(b-b_+\right)+\delta\left(b-b_-\right)\right)$$
Note that the $b-$integral is nonzero only if $b_{\pm}$ are real, which implies $|a-c|\leq S$. Thus, the desired integral now reads

$$\begin{align} P(S)&=\frac{S}{4}\iint_{|a-c|\leq S}\kern-2em\mathrm{d}a\,\mathrm{d}c~~e^{-(a^2+c^2)}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}b~\frac{e^{-\lambda b^2}}{|b|}\left(\delta\left(b-b_+\right)+\delta\left(b-b_-\right)\right)\\ &=S\,e^{-\frac{\lambda}{4}S^2}\iint_{|a-c|\leq S}\kern-2em\mathrm{d}a\,\mathrm{d}c~~\frac{e^{-(a^2+c^2)+\frac{\lambda}{4}(a-c)^2}}{\sqrt{S^2-(a-c)^2}} \end{align}$$
Introducing new variables $x=a-c$ and $y=a+c$,
$$\mathrm{d}a\,\mathrm{d}c=\left|\frac{\partial(a,c)}{\partial(x,y)}\right|\mathrm{d}x\,\mathrm{d}y=\frac{1}{2}\mathrm{d}x\,\mathrm{d}y,$$
we have
$$\begin{align} P(S)&=\frac{S}{2}e^{-\frac{\lambda}{4}S^2}\iint_{|x|\leq S}\mathrm{d}x\,\mathrm{d}y~\frac{e^{-\frac{y^2}{2}-\left(\frac{1}{2}-\frac{\lambda}{2}\right)x^2}}{\sqrt{S^2-x^2}}\\ &=\frac{S}{2}e^{-\frac{\lambda}{4}S^2}\color{Blue}{\int_{-\infty}^{\infty}\!\!\!\mathrm{d}y~e^{-\frac{y^2}{2}}}\int_{-S}^{S}\!\!\!\mathrm{d}x~\frac{e^{-\left(\frac{1}{2}-\frac{\lambda}{4}\right)x^2}}{\sqrt{S^2-x^2}}\\ &=\frac{S}{2}e^{-\frac{\lambda}{4}S^2}\color{Blue}{\sqrt{2\pi}}\cdot2\int_0^S\!\!\!\mathrm{d}x~\frac{e^{-\left(\frac{1}{2}-\frac{\lambda}{4}\right)x^2}}{\sqrt{S^2-x^2}}\\ &=\sqrt{2\pi}S\,e^{-\frac{\lambda}{4}S^2}\int_0^S\!\!\!\mathrm{d}x~\frac{e^{-\Lambda\left(x/S\right)^2}}{\sqrt{S^2-x^2}},\qquad\Lambda\equiv\left(\frac{1}{2}-\frac{\lambda}{4}\right)S^2\\ \end{align}$$
setting $x=S\sin\left(\frac{\theta}{2}\right)$, we have
$$\begin{align} P(S)&=\sqrt{\frac{\pi}{2}}S\,e^{-\frac{\lambda}{4}S^2}\int_0^{\pi}\!\!\!\mathrm{d}\theta~e^{-\Lambda\sin^2\left(\frac{\theta}{2}\right)}\\ &=\sqrt{\frac{\pi}{2}}S\,e^{-\frac{\lambda}{4}S^2-\frac{\Lambda}{2}}\int_0^{\pi}\!\!\!\mathrm{d}\theta~e^{\frac{\Lambda}{2}\cos\theta}\\ &=\sqrt{\frac{\pi}{2}}S\,e^{-\frac{\lambda}{4}S^2-\frac{\Lambda}{2}}\cdot\pi\,I_0\!\left(\frac{\Lambda}{2}\right), \end{align}$$
where $I_0(x)$ is the modified Bessel function of the first kind (see Eq. (5) of http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html). With some further simplifications, we have

$$\boxed{P(S)=\sqrt{\frac{\pi^3}{\!2}}S\exp{\!\left(-\frac{2+\lambda}{8}S^2\right)}\,I_0\!\left(\frac{2-\lambda}{8}S^2\right)}$$
Cheers!