1. ## integrals

1) $\int_{0}^{\infty} e^{-ax^{2}} \sin \Big( \frac{b}{x^{2}} \Big) \ dx$

2) $\int_{0}^{\infty} e^{-ax^{2}} \cos \Big(\frac{b}{x^{2}} \Big) \ dx$

3) $\int_{0}^{\infty} \cos ax^{2} \cos 2bx \ dx$

4) $\int_{0}^{\infty} \sin ax^{2}} \cos 2bx \ dx$

I can evaluate all of these integrals, but using a method that's just too difficult to justify. Namely evaluating an integral with real-valued parameters and assuming the solution is valid when one of the parameters is imaginary.

2. ## Re: integrals

Originally Posted by Random Variable
1) $\int_{0}^{\infty} e^{-ax^{2}} \sin \Big( \frac{b}{x^{2}} \Big) \ dx$
I am out of my depth here, but have an idea.

The solution is clearly real.

What if we consider the complex integral:
$\int_{0}^{\infty} e^{-az^{2}} \sin \Big( \frac{b}{z^{2}} \Big) \ dx$; where I assume a,b are real and z = x + iy

Now
$Re(\sin \Big( \frac{b}{z^{2}} \Big)) = 0$ when x = y.

also
$Re(\sin \Big( \frac{b}{z^{2}} \Big))$ tends to zero, when x is sufficiently large

So if we integrate along the path from the origin out to (R + iR) and then around the arc (centred on the origin) from R+iR to $\sqrt2R$ we find, after taking R to the limit of infinity, that the solution is zero.