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?

  • $\begingroup$ @HarryPeter Could you perhaps stop re-tagging questions for a wee while - editing questions brings them to the front page, and now quite a few of the front-page questions are simply re-tagged questions. Mass-retagging is best done at quiet times (e.g. weekends). Thanks :-) $\endgroup$ – user1729 Feb 10 '14 at 10:40

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 $f(0)=0$.

  • $\begingroup$ You mean $f$ is strictly decreasing on the real line? $\endgroup$ – Keith Aug 25 '19 at 19:58
  • $\begingroup$ $f$ is not necessarily decreasing. But $f(x)<0$, for $x>0$ and $f(x)>0$, for $x<0$. $\endgroup$ – Yiorgos S. Smyrlis Aug 25 '19 at 20:07
  • $\begingroup$ you said $e^acos(\xi) < 0$ for all $\xi$. I think that implies $f$ is strictly decreasing on the real line. $\endgroup$ – Keith Aug 25 '19 at 20:10

Partial solution. Since $f$ is odd, it suffices to consider only when $z > 0$ and we assume so. Let $\beta = -\alpha > 0$. Integrating by parts,

\begin{align*} f(z) &= \int_{0}^{1} e^{-\beta t^{2}} \sin (zt) \, dt \\ &= \left[ e^{-\beta t^{2}} \left( \frac{1-\cos(zt)}{z} \right) \right]_{0}^{1} + 2\beta\int_{0}^{1} t e^{-\beta t^{2}} \left( \frac{1-\cos(zt)}{z} \right) \, dt \\ &= e^{-\beta} \left( \frac{1-\cos z}{z} \right) + 2\beta \int_{0}^{1} t e^{-\beta t^{2}} \left( \frac{1-\cos(zt)}{z} \right) \, dt. \tag{1} \end{align*}

Since $1-\cos x \geq 0$ with equality only on a discrete set, we always have

$$\int_{0}^{1} t e^{-\beta t^{2}} \left( \frac{1-\cos(zt)}{z} \right) \, dt > 0$$

for all $z > 0$. This proves that $f(z) > 0$ for $z>0$ and that it has the unique zero $z = 0$.

Further remark. One may further be interested in the asymptotic behavior of $f$. Rearranging $\text{(1)}$, we have

$$ f(z) = \frac{1 - e^{-\beta}\cos z}{z} - \frac{2\beta}{z} \int_{0}^{1} t e^{-\beta t^{2}} \cos (zt) \, dt. $$

Utilizing the Riemann-Lebesgue lemma we know that $\int_{0}^{1} t e^{-\beta t^{2}} \cos (zt) \, dt = o(1)$ as $|z| \to \infty$. This shows

$$ f(z) = \frac{1 - e^{-\beta}\cos z}{z} + o\left( \frac{1}{z} \right). $$

(This error bound can be improved, but we do not pursue this direction.) We can also check this from the following graph:

enter image description here


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