Convergence of $f(t)= \frac{\exp(it)}{t^a}$

Question

Let $$f(t)= \frac{\exp(it)}{t^a}.$$

For what values ​​of $a$ does the integral $\displaystyle\int_{0}^{+\infty}f(t)dt$ converge?

• For $a>0$ it's clear with an integration by parts

Unfortunately in the case $a<0 $ I was not managed to prove (or not) the convergence.

I tried an IPP, equivalent, domination without success.

Any ideas?

Thanks

Answer 1

Presumably, $a\in \mathbb{R}$. Clearly if $a\leq0$, the integral diverges as $\int_{2n\pi}^{(2n+1)\pi}\dfrac{\sin t}{t^a}dt$ doesn't go to $0$. If $a\geq1$ The integral again diverges as $\int_0^1\frac{\cos x}{x^a}dx$ diverges.

What remains is $1>a>0$ In this case, split the region into intervals of size $\pi$ Taking the imaginary part, we have, $$\int_0^\infty\dfrac{\sin x}{x^a}dx=\sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\dfrac{\sin x}{x^a}$$ You can check that this forms an alternating series with decreasing terms. Hence it converges. Similarly, the integral with cos also converges, and hence the whole thing converges.