# On an asymptotic integral

Let $$\phi, a \in C^{\infty}([0,1])$$ and assume $$a(0)=1$$. Suppose that $$\int_0^1 e^{\tau \,\phi(t)}\,a(t)\,dt =0 \qquad \text{for all \tau \in \mathbb R}.$$ Does it follow that $$\phi$$ is a constant in a neighborhood of $$t=0$$?

If the answer to the above question is affirmative I have a follow up question as follows: Suppose additionally that $$b \in C^{\infty}([0,1]$$ with $$b(0)=1$$ and instead of the latter equation there holds $$\Bigg |\int_0^1 e^{\tau \,\phi(t)}\,a(t)\,dt\Bigg| \leq \frac{1}{|\tau|} \Bigg|\int_0^1 e^{\tau \,\phi(t)}\,b(t)\,dt\Bigg|$$ Does it again follow that $$\phi(t)$$ is constant in a neighborhood of $$t=0$$?

• no; just try $\phi(t)=\text{constant}\neq 0$ and take for $a(t)$ any function with $\int_0^1 a(t)dt=0$; then obviously also $e^\tau \int_0^1 a(t)dt=0$ for any $\tau\in\mathbb{R}$. Mar 14, 2022 at 16:44
• This was a bad typo on my part. I meant to write derivative of $\phi$ in the question. sorry about this.
– Ali
Mar 14, 2022 at 17:00

No. E.g., let $$\phi(t)=(t-1/2)^2$$ (so that $$\phi$$ is even about $$1/2$$) and $$a(t)=2(1/2-t)$$ (so that $$a$$ is odd about $$1/2$$).