# 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

• Identical question also posted on physics.SE. – Mike Jan 8 '18 at 20:45
• @Mike This looks likely to get closed on Physics SE as there's no physics content as such, so it might be best to leave it on Maths SE if it's within the rules here. – StephenG Jan 8 '18 at 21:11
• @StephenG I have no problem with that. I just feel that it's relevant information. – Mike Jan 8 '18 at 21:21
• Use Eq. (7) of mathworld.wolfram.com/DeltaFunction.html, and resolve the delta function – Siddhant Das Jan 9 '18 at 1:00
• @Mike, Stephen, Removed from Ph SE. – math Jan 9 '18 at 3:37

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!