I have an integral of the form $$ I = \int\limits^{1}_{0} \exp\left(\dfrac{vt}{(v+1)^2 + v^2} - vt\right) dv $$ and I want to prove that $I\leq c t^{-1}$ for the sufficiently large $t$, where $c$ is a constant independent of $t$. Can anyone give me some hints or references to prove this expansion?
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4$\begingroup$ The integrand is less than $\exp(-2tv^2)$. $\endgroup$– Brendan McKayCommented Aug 16, 2022 at 6:34
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2$\begingroup$ The bound $\exp(-2tv^2)$ gives $O(t^{-1/2})$ for the integral, not $O(t^{-1})$. And indeed plotting the value of the integral strongly suggests an asymptotic value about $0.63 t^{-1/2}$. So I believe the claim is false. $\endgroup$– Brendan McKayCommented Aug 16, 2022 at 9:34
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1$\begingroup$ A lower bound is $\exp(-\frac45 t v^2)$. This completes the proof that the value is $\Theta(t^{-1/2})$. $\endgroup$– Brendan McKayCommented Aug 16, 2022 at 9:51
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1$\begingroup$ @BrendanMcKay Upper and lower bounds look interchanged (without affecting the resut). $\endgroup$– Giorgio MetafuneCommented Aug 16, 2022 at 10:27
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1$\begingroup$ @GiorgioMetafune Quite right, thanks. Upper and lower are interchanged. Also, for large $t$ the lower bound is sharper and that suggest the asymptotic value is $\sqrt{\pi/8t}$ which matches experiment. $\endgroup$– Brendan McKayCommented Aug 16, 2022 at 12:38
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2 Answers
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(This question should be on math.stackexchage.com.)
Substitute $v=t^{-1/2}u$, then it becomes $$ t^{-1/2} \int_0^{t^{1/2}} e^{-2u^2}\bigl(1 + O(u^3/t^{1/2})\bigr)\,du = \sqrt{\frac{\pi}{8t}} + O(t^{-1}).$$
answered Aug 16, 2022 at 13:11
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1$\begingroup$ Answering questions that belong on MSE here on MO doesn't help to convince people to put them where they belong, though. If one must answer, then I think that it is better to do it as a comment. Either way, it is probably a good idea for you to vote to closre or migrate. $\endgroup$– LSpiceCommented Aug 16, 2022 at 17:22
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1$\begingroup$ @LSpice You are quite right and I voted to migrate. $\endgroup$ Commented Aug 17, 2022 at 2:53
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Suppose that is exists a $k$ such that
$$\dfrac{vt}{(v+1)^2 + v^2} - vt=-\frac{2 v^2 (v+1)}{2 v^2+2 v+1}t\sim -k\,v^2\,t$$
The integral would be $$I\sim \frac{\sqrt{\pi }}{2} \frac{\text{erf}\left(\sqrt{kt}\right)}{\sqrt{kt}}$$ Expanded for large values of $t$
$$I \sim\frac{\sqrt{\pi }}{2}\left(\frac{1}{\sqrt{k t}} -\frac{e^{-k t}}{\sqrt{\pi } k t} \right)$$
Such a $k$ exists and its value is $\sim \frac 53$
