# Estimate for a simple oscillatory integral

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.

• 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 $\lambda^{-1/2}$. Dec 1 '20 at 2:18
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 $$t\to\infty$$.
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 $$\int_{-1}^1e^{i\lambda u^2}\,du=\int_{-\infty}^{\infty}e^{i\lambda u^2}du+\mathcal{o}(\lambda^{-1/2})=\sqrt{\pi}e^{i\pi/4}\lambda^{-1/2}+\mathcal{o}(\lambda^{-1/2}).$$