0
$\begingroup$

PROBLEM

I am trying to prove the following identity:

$$ \int_{-\infty}^{+\infty} dx \frac{e^{ibx}-e^{iax}}{x} = 2\pi i , \hspace{5mm} b>0, a<0$$

This is just a conjecture that I made while trying to complete a proof for Fourier's Inversion Theorem. Numerically it seems to be correct; moreover, assuming that the integral converges, I proved that it is indeed correct by contour integration of the corresponding Cauchy's Principal Value.

ATTEMPT

As stated above, I just need to prove convergence. Intuitively it makes sense because there will be oscillations just as in the integral $\int_{-\infty}^{+\infty}dx \sin(x)/x$ but this time there is the problem that I would have to bring both the cosines out together when using Euler's Formula in order to avoid divergence in $x=0$. I haven't tried this way yet but if necessary I could approach it. On the other hand I tried using the exponential power series definition and I ended up with:

$$ \sum_{n=0}^{+\infty} \frac{(i)^{n+1}}{(n+1)!(n+1)}(\beta^{n+1}-\alpha^{n+1})(b^{n+1}-a^{n+1}) $$

where $\alpha,\beta$ are the lower and upper limits of the integral respectively (so that they must be sent to $+\infty$ and to $-\infty$ separately). However I do not see yet how this could be helpful.

Any answer or comment is welcome and let me know if I cam explain myself clearer!

$\endgroup$
4
  • $\begingroup$ Does the improper integral converge? If not, in what sense do you want to interpret it? $\endgroup$
    – GEdgar
    Jul 9, 2023 at 20:14
  • $\begingroup$ I'm not sure I understand. If you proved the above equality is correct by contour integration, didn't you inherently prove the integral converges because you got it to equal something? $\endgroup$ Jul 9, 2023 at 20:26
  • $\begingroup$ @Accelerator Unfortunately that is not enough since I used contour integrals to evaluate the Cauchy’s Principal Value and it may happen that while it converges the integral itself doesn’t. $\endgroup$ Jul 9, 2023 at 20:46
  • $\begingroup$ Oh, that's what you meant. Thank you for clarifying. $\endgroup$ Jul 9, 2023 at 20:54

1 Answer 1

3
$\begingroup$

OP's integral identity easily follows from Dirichlet integral:

$$\mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{e^{i\xi x}}{x} \, \mathrm{d}x = i \int_{-\infty}^{\infty} \frac{\sin \xi x}{x} \, \mathrm{d}x = i\pi\operatorname{sgn}(\xi), \qquad \xi \in \mathbb{R}$$

To show the convergence of the integral, we may take advantage of the oscillatory behavior of the imaginary exponential and perform integration by parts:

\begin{align*} &\int_{p}^{q} \frac{e^{ibx}-e^{iax}}{x} \, \mathrm{d}x \\ &= \left[ \frac{1}{x}\left(\frac{e^{ibx}-1}{ib}-\frac{e^{iax}-1}{ia}\right) \right]_{p}^{q} + \int_{p}^{q} \frac{1}{x^2}\left(\frac{e^{ibx}-1}{ib}-\frac{e^{iax}-1}{ia}\right) \, \mathrm{d}x \\ &\to \int_{-\infty}^{\infty} \frac{1}{x^2}\left(\frac{e^{ibx}-1}{ib}-\frac{e^{iax}-1}{ia}\right) \, \mathrm{d}x \qquad \text{as} \quad p \to -\infty \text{ and } q \to +\infty. \end{align*}

The last integral converges absolutely, since the integrand is continuous (in fact, analytic) and decays at a speed of $\mathcal{O}(x^{-2})$ as $x\to\pm\infty$.

Remark. This is a special case of Dirichlet test for integrals.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .