4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @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
3
$\begingroup$

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 $f(0)=0$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.