Inequality involving Gaussian integral [closed]

Question

I'm looking to prove the following inequality:

$$ \left| \int_0^1 e^{-x^2} \sin(x) \, dx \right| \leq \frac{1}{2} \left(1- \frac{1}{e}\right) $$

So far I have no idea on how to prove it. Anybody?

Answer 1

Writing $\sin x=\frac1{2i}(e^{ix}-e^{-ix})$ and then $e^{-x^2}e^{\pm ix}=e^{-1/4}e^{-(x\mp i/2)^2}$, we get $$\int_0^1 e^{-x^2}\sin x\,dx=\frac{\sqrt{\pi } \left(i\, \text{erf}\left(1+\frac{i}{2}\right)+2 \text{erfi}\left(\frac{1}{2}\right)-\text{erfi}\left(\frac{1}{2}+i\right)\right)}{4 e^{1/4}},$$ which is $0.294\ldots$; here $\text{erf}$ and $\text{erfi}$ denote, as usual, the error function and the imaginary error function, respectively.

On the other hand, $\frac12(1-\frac1e)=0.316\ldots>0.294\ldots$. This establishes your inequality.