Define $$f(z)=\int_0^1\mathrm{e}^{\alpha t^2}\sin(tz)\,dt,$$ where $\alpha \in \mathbb{R}$. If $\alpha >0$ then $f(z)$ has infinitely many real zeros and at most a finite number of complex zeros. What if $\alpha <0$?
Hint. Integrate by parts.
My attempt so far. To get some intuition about the problem, I tried to demonstrate the claim in the question. If $x\in \mathbb{R}$, it seems that $$f(x) = 0 \iff \int_0^1\mathrm{e}^{\alpha t^2 + tx}dt= \int_0^1\mathrm{e}^{\alpha t^2 - tx}dt.$$
I was unable to simplify that condition. I then attempted to integrate by parts blindly, and turn the $\alpha <0$ case into the $\alpha>0$ case, and use the claim. $$\int_0^1 \mathrm{e}^{\alpha t^2}\sin(tz)\,dt = -\frac1z \mathrm{e}^\alpha \cos(z) + \frac1z + \frac1z \int_0^1 \cos(tz)\,\mathrm{e}^{\alpha t^2}2\alpha t \, dt.$$
And there I got stuck. Any ideas?
This is not a complete solution. It provides however an alternative expression of $f$, which allows us to obtain that $f$ vanishes only at one real value ($z=0$), if $a<0$.
We have that \begin{align*} f'(z) &=\int_0^1 \mathrm{e}^{a t^2}t\cos (tz)\,dt=\left.\frac{1}{2a}\mathrm{e}^{at^2}\cos(tz)\right|_{t=0}^{t=1}+\frac{z}{2a}\int_0^1 \mathrm{e}^{at^2}\sin(tz)\,dz \\ &= \frac{1}{2a}\mathrm{e}^a\cos z-\frac{1}{2a}+\frac{z}{2a} f(z). \end{align*} Thus $$ \mathrm{e}^{-z^2/4a}\left(f'(z)-\frac{z}{2a}f(z)\right)=\mathrm{e}^{-z^2/4a} \left(\frac{1}{2a}\mathrm{e}^a\cos z-\frac{1}{2a}\right) $$ or $$ \left(\mathrm{e}^{-z^2/4a}f(z)\right)^{\!\prime}=\mathrm{e}^{-z^2/4a} \left(\frac{1}{2a}\mathrm{e}^a\cos z-\frac{1}{2a}\right). $$ Thus $$ f(z)=\frac{\mathrm{e}^{z^2/4a}}{2a}\int_0^z \mathrm{e}^{-\zeta^2/4a} \left(\mathrm{e}^a\cos \zeta-1\right)\,d\zeta. $$ Now in this expression we observe the following thing: If $a<0$, then $\mathrm{e}^a\cos \zeta-1<0$, for all $\zeta$, and thus $f$ is strictly increasing on the real line, which means that it has a unique zero, namely $f(0)=0$.