# Upper bound of an oscillatory integral

Suppose that $$\mathrm{f}:[\mathrm{a}, \mathrm{b}] \rightarrow \mathbb{R}$$ is a smooth, convex function, and there exists a constant $$\mathrm{t}>0$$ such that $$f^{\prime}(x) \geq t$$ for all $$x \in(a, b)$$. Prove that $$\left|\int_a^b e^{i f(x)} d x\right| \leq \frac{2}{t}$$

I tried to use some of the properties from the oscillatory integral and wrote the integrand as $$f^{\prime}(x) e^{i f(x)} / f^{\prime}(x)$$ but it seems there is some fundamental idea missing in my attempts.

• try writing the integrand as $f'(x)e^{if(x)}/f'(x)$ and see if it helps Apr 28, 2023 at 11:06
• @tomos Thank you for the suggestion. That's also what I did, my bad for not mentioning it in the post. Unfortunately, it didn't lead to any fruitful results. Apr 28, 2023 at 11:21
• Notice that \begin{align} e^{i f(x)} = \frac{1}{i f'(x)}\frac{d}{d x}e^{i f(x)} \end{align} which means \begin{align} \int^b_a e^{i f(x)} d x = \int^b_a \frac{1}{i f'(x)}\frac{d}{d x}e^{i f(x)} d x. \end{align} Then integration by parts. Finally, estimate your way to success. Apr 28, 2023 at 11:33
• @JackyChong Thanks for the idea, but integration by parts doesn't lead to results as far as I try because of the second derivative of $f(x)$ that comes to play. Could you please elaborate more? Apr 28, 2023 at 12:17

Using integration by parts and noting that $$f'(x) \geq t > 0$$ and $$f''(x) \geq 0$$,
\begin{align*} \left| \int_{a}^{b} e^{if(x)} \, \mathrm{d}x \right| &= \left| \frac{e^{if(b)}}{if'(b)} - \frac{e^{if(a)}}{if'(a)} + \int_{a}^{b} \frac{f''(x)}{if'(x)^2}e^{if(x)} \, \mathrm{d}x \right| \\ &\leq \frac{1}{f'(b)} + \frac{1}{f'(a)} + \int_{a}^{b} \frac{f''(x)}{f'(x)^2} \, \mathrm{d}x \\ &\leq \frac{1}{f'(b)} + \frac{1}{f'(a)} + \left[ \frac{1}{f'(a)} - \frac{1}{f'(b)} \right] \\ &= \frac{2}{f'(a)} \leq \frac{2}{t}. \end{align*}
Remark. This bound cannot be improved further by noting that the equality holds if $$f$$ is linear with slope $$t$$ and $$t(b - a)$$ is an odd multiple of $$\pi$$.
• Thank you for the solution. I wanted to ask how you get $e^{if(x)}$ is less than $i$ in modulus expression. It seems you have applied this rule to every part of the triangle inequality so I was wondering how it could be proved. Apr 28, 2023 at 13:54
• @Snowball No problem. Note that $|e^{iy}|=1$ for any real $y$ as well as $|i|=1$. Now you mix this with the triangle inequality. Apr 28, 2023 at 13:55