Monday, March 7, 2016

calculus - Gaussian type integral intinftyinftyfracmathrmea2x21+x2mathrmdx




When working a proof, I reached an expression similar to this:



ea2x21+x2dx



I've tried the following:



1. I tried squaring and combining and converting to polar coordinates, like one would solve a standard Gaussian. However, this yielded something which seems no more amenable to a solution:



θ=2πθ=00rea2r2(1+r2sin2(θ))(1+r2cos2(θ))drdθ




2. I tried doing a trig substitution, t = tan u, and I have no idea what to do from there.



π2π2ea2tan2(u)du



3. I looked into doing u2=1+x2 but this gives us a ugly dx that I don't know how to handle, and moreover, I think I'm breaking my limits of integration (because Mathematica no longer solves it.):



u2=1+x2



2udu=2xdx




dx=uu21
ea2ea2u2uu21du



4. I looked into some form of differentiation under the integral, but that didn't seem to yield anything that looked promising. (I checked parameterizing x^2 to x^b in both places, and in either place, and nothing canceled cleanly.)



I have a solution from Mathematica, it's:



πea2erfc(a)



But I'd like to know how to arrive at this. I'm sure it's something simple I'm missing.



Answer



Let F be the function
F(a)=ea2x21+x2dx
We take the derivative w.r.t a
F(a)=dda(ea2x21+x2dx)=dda(ea2x21+x2)dx=2ax2ea2x21+x2dx
=2a((x2+1)1)ea2x21+x2dx=2aea2x2dx+2aF(a)=2aπa2+2aF(a)
Then

F(a)=2a(F(a)π1|a|)=2aF(a)2πsign(a).
Then you have a differential equation:
F(a)2aF(a)=2πsign(a)
with initial condition F(0)=π.
This fisrt order ode has integrant factor:
μ(a)=e2ada=ea2
Then
(ea2F(a))=2πsign(a)ea2

this implies
ea2F(a)=2πsign(a)ea2da+C
Finaly
F(a)=ea2(C2πsign(a)ea2da)


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