Consider the following integral:
$$\int_{0}^{\infty}f(ix)dx$$
And consider a scenario where the following u-sub would be appropriate to solve the integral:
$u = ix, du=idx$
$$=\int_{0}^{\infty}\frac{1}{i}f(u)du$$
Why is this u-substitution allowed? I understand how you can change the lower bound, because $0\times i=0.$ However…. leaving the upper bound at $\infty$ is bothersome to me. It should intuitively make sense that $\infty$ times any positive number is $\infty$. Maybe we can… just extend this rule to complex numbers? I'm aware of an example in which this is used, but I have NO idea the justification. You're really making the claim that $\infty \times i=\infty.$ Can we always do this? What if our complex number was $-1-i$?
Edit: To clarify, I will provide the example. The integral is:
$$\int_{0}^{\infty} e^{-x^2} \cos (x^2)dx$$
Transform it as follows:
$$\text{Re}\int_{0}^{\infty}e^{(-1+i)x^2}dx$$
The substitution is $u=\sqrt{-1+i}x$ to transform this into a Gaussian integral. How do we justify the bounds from this point being $0$ and $\infty$? Are we saying that $\sqrt{-1+i}\times\infty = \infty$? Does the fact that we're only taking the real part have anything to do with this?
Best Answer
If $I$ is given by
$$I=\int_0^\infty f(ix)\,dx$$
then the substitution $x\to -i y$ gives
$$I=-i \int_0^{i\infty}f(y)\,dy$$
It is not true, in general, that $I=-i \int_0^{\infty}f(y)\,dy$.
If $f(z)$ is, however, analytic in the first quadrant of the $z$-plane, and decays sufficiently rapidly such that
$$\lim_{R\to \infty}\int_0^{\pi/2}f(Re^{i\phi})\,iRe^{i\phi}\,d\phi=0$$
then Cauchy's Integral Theorem guarantees that
$$\int_0^{iR} f(x)\,dx+\int_{\pi /2}^0 f(Re^{i\phi})\,iRe^{i\phi}\,d\phi+\int_{R}^0 f(y)\,dy=0$$
which after letting $R\to\infty$, we find that
$$\int_0^{i\infty} f(x)\,dx=\int_0^\infty f(y)\,dy$$