1. ## related integrals

$\int_{0}^{\infty} \cos \Big(x + \frac{1}{x} \Big) \frac{dx}{\sqrt{x}} = 2 \int^{\infty}_{0} \cos \Big(x^{2}+ \frac{1}{x^{2}} \Big) dx = 2 \ \Re \int^{\infty}_{0} e^{-i(x^{2}+\frac{1}{x^{2}})} dx$

and $\int_{0}^{\infty} \sin \Big(x + \frac{1}{x} \Big) \frac{dx}{\sqrt{x}} = 2 \int^{\infty}_{0} \sin \Big(x^{2}+ \frac{1}{x^{2}} \Big) dx = -2 \ \Im \int^{\infty}_{0} e^{-i(x^{2}+\frac{1}{x^{2}})} dx$

Could I use $\int^{\infty}_{0} e^{-a(x^{2}+\frac{1}{x^{2}})} dx = \frac{\sqrt{\frac{\pi}{a}}}{2e^{2a}}$ and substitute i for a?

2. ## Re: related integrals

Good question! Any kind of substitution would have to be justified. It boils down to whether the last result extends to complex $a$'s. I'll need to think about this a while.

3. ## Re: related integrals

The answer to your question is yes you can. You can extend the identity to complex values of $a$, in a suitable neighborhood, because of the uniqueness principle in complex analysis.

4. ## Re: related integrals

I'm not familiar with a uniqueness principle. All I know is that when $a$ is real valued and $a \le 0$, the integral diverges. I don't know for what imaginary or complex values the integral is convergent. Substituting $a= i$ does lead to the correct solution, though.

EDIT: It should be OK if I can show that $\int^{\infty}_{0} \cos \Big( x^{2}+\frac{1}{x^{2}} \Big) \ dx$ and $\int^{\infty}_{0} \sin \Big( x^{2} + \frac{1}{x^{2}} \Big) \ dx$ actually converge.

5. ## Re: related integrals

The question of whether an identity involving a real parameter can extended to complex values comes up a lot in complex analysis. There's more to it than just showing that certain integrals converge. A reference for the uniqueness principle is Gamelin's book on complex analysis. He addresses questions very similiar to the one you're considering.