Change of variable x=iy in improper integral

Question

I'm trying to solve the following question:

Let $I=\int_0^{\infty}\exp(-x^4)dx$. Take $x=i y$ to get $i\int_0^{\infty}\exp(-y^4)dy=i I$. Explain.

This change of variable implies $I=iI$, with $I$ real, so $I=0$, which is wrong. I have to explain why this change of the integral does not work. Does anyone know how to handle this problem?

Answer 1

What you have found, rather, is that

$$\lim_{R\to\infty} \int_0^{i R} dy \, e^{-y^4} = -i \int_0^{\infty} dx \, e^{-x^4}$$

Answer 2

Why not something more "pedestrian"?:

$$u=x^4\implies dx=\frac{du}{4 u^{3/4}}\implies \int\limits_0^\infty e^{-x^4}dx=\frac14\int\limits_0^\infty u^{-3/4}e^{-u}du=\frac14\Gamma\left(\frac14\right)=\Gamma\left(\frac54\right)$$