Change of variable leads to contradiction for an elementary integral.

Question

Integral $I_1(\alpha,\beta)=\int_0^\infty t^\alpha \exp(-i t^{\beta}) ~dt$ converges for $-1<\alpha<\beta-1$. By introducting $u = t^\beta$ the integral is reduced to $\frac{1}{\beta}\int_0^\infty u^{\frac{\alpha+1}{\beta}-1} \exp(-i u)~du$ and the condition for the convergence is easily obtained. However, another change of variable by $t=1/\tau$ yields $I_2(\alpha,\beta)=\int_0^\infty \tau^{-\alpha-2} \exp(-i \tau^{-\beta}) ~d\tau$. Thus we have $I_2(\alpha,\beta) = I_1(-\alpha-2,-\beta)$. This implies the condition for the convergence is $-1<-\alpha-2<-\beta-1$. By rearranging terms, we get $\beta-1<\alpha<-1$. This range is totally dijoint with the original condition $-1<\alpha<\beta-1$. This is obviously a contradiction. What did I do wrong?

Answer 1

You can't plug $-\beta$ into your formula in place of $\beta$, since your formula only works for $\beta\gt0$. Working with a negative $\beta$ inside the exponential gives different results.

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If $\beta\gt0$, then substituting $t\mapsto t^{1/\beta}$ gives $$ \begin{align} \int_0^\infty t^\alpha\exp\!\left(it^\beta\right)\,\mathrm{d}t &=\frac1\beta\int_0^\infty t^{\frac{\alpha+1}\beta-1}\exp(it)\,\mathrm{d}t\tag{1}\\ &=\frac1\beta\,\exp\!\left(i\frac\pi2\frac{\alpha+1}\beta\right)\Gamma\!\left(\frac{\alpha+1}\beta\right)\tag{2} \end{align} $$ which converges when $0\lt\frac{\alpha+1}\beta\lt1$ which equivalent to $0\lt\alpha+1\lt\beta$.

However, if $\beta\lt0$, then $\exp\,\left(it^\beta\right)$ behaves quite differently. It doesn't oscillate near $\infty$ and oscillates quite quickly near $0$. $$ \begin{align} \int_0^\infty t^\alpha\exp\!\left(it^\beta\right)\,\mathrm{d}t &=\int_0^\infty t^{-\alpha-2}\exp\!\left(it^{-\beta}\right)\,\mathrm{d}t\tag{3}\\ &=-\frac1\beta\int_0^\infty t^{\frac{\alpha+1}\beta-1}\exp(it)\,\mathrm{d}t\tag{4}\\ &=-\frac1\beta\,\exp\!\left(i\frac\pi2\frac{\alpha+1}\beta\right)\Gamma\!\left(\frac{\alpha+1}\beta\right)\tag{5} \end{align} $$ which converges for $0\lt\frac{\alpha+1}\beta\lt1$ which is equivalent to $\beta\lt\alpha+1\lt0$.