Estimate for a simple oscillatory integral

Question

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!

Answer 1

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 $t\to\infty$.

Answer 2

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 $$\int_{-1}^1e^{i\lambda u^2}\,du=\int_{-\infty}^{\infty}e^{i\lambda u^2}du+\mathcal{o}(\lambda^{-1/2})=\sqrt{\pi}e^{i\pi/4}\lambda^{-1/2}+\mathcal{o}(\lambda^{-1/2}).$$