How can I justify the substitution of a complex variable in the following integral?

Question

$ \displaystyle I(k) = \int_{0}^{\infty} e^{(i-1)x}x^{k} \mathrm{d}x \tag*{} $

I want to know how can we justify the substitution $ (1-i)x=t $ while the upper and lower limits of the integral remains the same.

Answer 1

This is done using Cauchy integral theorem. For any $0<r<R$, by the theorem, $$\left[\int_r^R+\int_R^{(1-i)R}+\int_{(1-i)R}^{(1-i)r}+\int_{(1-i)r}^r\right]z^k e^{-z}\,dz=0.$$ Here we assume that $\Re k>-1$, and the principal value of $z^k$ is taken.

Now let $r\to 0$ and $R\to\infty$. The 2nd and 4th integrals vanish. The 3rd one, after substituting $z=(1-i)x$, is $-(1-i)^{k+1}\int_r^R x^k e^{(i-1)x}\,dx$ and tends to $-(1-i)^{k+1}I(k)$. Finally, the 1st one tends to $\int_0^\infty z^k e^{-z}\,dz=\Gamma(k+1)$.

Answer 2

$$\int_0^\infty e^{(i-1)x}x^k\,dx=\int_0^\infty (\cos x+i\sin x)e^{-x}x^k\,dx.$$

The antiderivative is of the form $(\cos(x)\,P(x)+\sin(x)\,Q(x))\,e^{-x}$ where $P,Q$ are polynomials. Hence the limit $x\to\infty$ is zero, and only the bound $x=0$ brings a contribution.