Divergence/convergence of an integral $\int_{-\infty}^{+\infty} (1-e^{ix}) |x|^{-n} dx$

Question

I am told that the following integral converges for $1<n<3$. $$ \int_{-\infty}^{+\infty} (1-e^{ix}) |x|^{-n} dx $$ I am a bit baffled. Anyone with a clue or where to start with this in order to evaluate it ?

Answer 1

Given that $1-e^{ix}$ is bounded, the "far" side convergence is given by the following:

$$\int^\infty (1-e^{ix})|x|^{-n}dx<2\int^\infty |x|^{-n}dx$$ which converges for $n>1$.

Around zero, you can expand the exponential: $1-e^{ix}\approx -ix+x^2/2$. Now write the improper integral

$$\int_{-\epsilon}^\epsilon (1-e^{ix})|x|^{-n}dx\approx \int_{-\epsilon}^\epsilon -ix |x|^{-n}dx+ \int_{-\epsilon}^\epsilon |x|^{2-n}dx$$ The first term has a zero principal value because the integrand is odd. The second term converges when $2-n>-1$ which is known behaviour of power function around zero.

Answer 2

For $|x|$ large $$ |1-e^{ix}|\,|x|^{-n}\le2\,|x|^{-n}. $$ The integral converges at $\pm\infty$ if $n>1$.

For $|x|$ small $|1-e^{ix}|\sim|x|$ and the integral converges at $0$ if $n<2$.

What happens if $2\le n<3$? The integral is not convergent at $x=0$. However, for any $r>0$ $$ \int_{|x|>r}(1-e^{ix})\,|x|^{-n}\,dx=\int_{|x|>r}(1-\cos x)\,|x|^{-n}\,dx-i\int_{|x|>r}\sin x\,|x|^{-n}\,dx. $$ The second integral is equal to $0$ (because $\sin x$ is odd). As for the first one we have $1-\cos x\sim x^2/2$ for $x$ close to $0$, and the integral converges if $n<3$. Thus, if $2\le n<3$ the integral id defined as a principal value.

Answer 3

I think the convergence of the integral is covered fine in the other answers, so just let me give you a non-rigorous (depending on your beliefs) way of calculating its value.

With the definition of Fourier transform $\hat f(\xi)=\int_{-\infty}^{+\infty}e^{-ix\xi}f(x)\,dx$ your integral can be seen to be $\hat{f}(0)-\hat{f}(-1)$ (if we agree to split the integral!), where $f(x)=|x|^{-n}$. It is known (see for example the book on classical Fourier analysis by Grafakos) that the Fourier transform of $f$ is $$ \hat{f}(\xi)=2|\xi|^{n-1}\Gamma(1-n)\sin(n\pi/2). $$ This gives that your integral equals $$ \hat{f}(0)-\hat{f}(-1)=0-2\Gamma(1-n)\sin(n\pi/2)=-2\Gamma(1-n)\sin(n\pi/2). $$