1
$\begingroup$

If $\varphi$ is a smooth function on $\mathbb{R}$, then integration by parts implies that there exists a constant $C>0$ such that $$ \Big|\int_0^1 \varphi(x)\, e^{i \lambda x}\, dx\Big|<\frac{C}\lambda $$ as $\lambda\rightarrow\infty$.

$\textbf{My question}$ is, whether one can determine the rate of decay, in terms of $\lambda$, of the oscillatory integral $$ \Big|\int_0^1 \frac{1}{\sqrt{x}}\, e^{i \lambda x}\, dx\Big|? $$ Since $\frac{1}{\sqrt{x}}\in L^1((0,1))$, it follows by the Riemann-Lebesgue Lemma that $\int_0^1 \frac{1}{\sqrt{x}}\, e^{i \lambda x}\, dx$ vanishes as $\lambda\rightarrow\infty$. However, Riemann-Lebesgue Lemma doesn't say anything about the rate of decay, hence my question.

Thanks for reading!

$\endgroup$
2
  • 2
    $\begingroup$ You should think of this as an alternating sum (consider the real part and estimate the contribution on pieces of size $\frac 1\lambda$). The first piece carries most of the weight, so the estimate should be $\lambda^{-1/2}$. $\endgroup$ Dec 1 '20 at 2:18
  • $\begingroup$ Interesting perspective, thanks for sharing! $\endgroup$
    – Tony419
    Dec 1 '20 at 4:14
3
$\begingroup$

By the substitution $tx=u$, the integral in question is $$\int_0^1\frac{e^{itx}}{\sqrt x}\,dx=\frac1{\sqrt t}\,\int_0^t\frac{e^{iu}}{\sqrt u}\,du \sim\frac1{\sqrt t}\,\int_0^\infty\frac{e^{iu}}{\sqrt u}\,du =(1+i) \sqrt{\frac{\pi }{2}}\frac1{\sqrt t}$$ as $t\to\infty$.

$\endgroup$
2
$\begingroup$

You can first substitute $u=\sqrt{x}$ so that your integral is $$\int_{-1}^1e^{i\lambda u^2}\,du.$$ As $\lambda\to\infty$, you can immediately see that most of the contribution from the integral will arise from roughly the region $[-1/\sqrt{\lambda}, 1/\sqrt{\lambda}]$, as beyond this interval the integrand is highly oscillatory, and the contributions cancel each other. So you expect that the integral is $\mathcal{O}\left(\lambda^{-1/2}\right)$.

For a more precise estimate, you can extend the region of integration to the entire real line as $\lambda\to\infty$, adding a small error $\mathcal{o}(\lambda^{-1/2})$, to obtain $$\int_{-1}^1e^{i\lambda u^2}\,du=\int_{-\infty}^{\infty}e^{i\lambda u^2}du+\mathcal{o}(\lambda^{-1/2})=\sqrt{\pi}e^{i\pi/4}\lambda^{-1/2}+\mathcal{o}(\lambda^{-1/2}).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.