3
$\begingroup$

I would like to show that the following function of the positive integers k and N does not grow to infinity with k, at least for some values of N. $$ F(k,N) = \int_1^{k^2} e^{\log (N/k) \sqrt x } e^{ x/k } \textrm{d}x $$ In other words, my goal is to find $N \in \mathbb{N}_{>0}$ and $M < \infty$ such that $f(k,N) < M$ for every $k$. How could I upper bound the integral efficiently? In the ideal case such a finite upper bound on $F(k,N)$ (uniform in $k$) will not grow faster than linearly with $N$...

$\endgroup$
3
  • 1
    $\begingroup$ Is $\log$ specifically the natural logarithm $\ln$ or any arbitrary logarithm of any base? $\endgroup$
    – FShrike
    Jul 29, 2021 at 9:59
  • 1
    $\begingroup$ Natural logarithm. However it does not make much difference... $\endgroup$ Jul 29, 2021 at 12:03
  • $\begingroup$ What have you tried for deriving an upper bound? What's the context of this integral? Please take a look at Enforcement of Quality Standards. $\endgroup$
    – Ѕааԁ
    Jul 29, 2021 at 13:42

3 Answers 3

5
+150
$\begingroup$

This is more of a supplement to Gary's answer, with the goal of avoiding the use of special functions.

There are two distinct behaviours depending on the size of $\log(N/k)$. Setting $m = \log(N/k)$, we have an integral of the form $$ \int_1^{k^2} e^{\frac{x +mk \sqrt{x}}{k}}.$$ Now notice that $(x +mk\sqrt{x}) < 0 \iff \sqrt{x} < -mk \iff x < m^2 k^2.$ So, in particular, if $m <- 1,$ then the integrand is upper bounded by $$ \int_1^{k^2} e^{(m+1)\sqrt{x} }\mathrm{d}x \le \int_0^\infty e^{(m+1)\sqrt{x}} \mathrm{d}x = \int_0^\infty \frac{2t}{(m+1)^2} e^{-t} \mathrm{d}t = \frac{2}{(m+1)^2},$$ where I've used that $\int _0^\infty te^{-t} = 1.$


If, on the other hand, $m > -1,$ then notice that the integrand is maximised near $k^2$, and grows quite large around there. This sets up the calculation $$ \int_1^{k^2} e^{\frac{x + mk\sqrt{x}}{k}} = \int_{1/k}^1 2k^2 t e^{k(t^2 +mt)} \le \int_0^1 2k^2 t e^{k(t^2 + mt)} \le e^{k(1+m)} \int_0^12k^2t = k^2 e^{k(1+m)}.$$

I'll note that for fixed $N/k$, the growth rate $k^2 e^{k(1+m)}$ is correct up to a factor of $k$. For a lower bound (for $k\ge 4$), observe that the slope of $t^2 + mt$ over $(0,1)$ is at most $2 + m$, and thus in $(1-1/k, 1),$ $t^2 + mt \ge 1 - (2+m)/k,$ giving $$ \int_{1/k}^1 2k^2 t e^{k(t^2 + mt)} \ge \int_{1 - 1/k}^1 2k^2 te^{k(t^2 + mt)} \ge \frac{2}{4e^{2}} ke^{-m} e^{k(1+m)} = \frac{k}{2e^2 (N/k)} e^{k(1+m)}$$

This quite clearly shows that $f(k,N)$ has a phase transition around the line $N/k = 1/e$ as $k$ grows large.

$\endgroup$
1
  • $\begingroup$ This is a much simpler approach! (+1) $\endgroup$
    – Gary
    Jul 30, 2021 at 11:25
4
$\begingroup$

Let $m=\log(N/k)$ and assume $k \gg N$. Then \begin{align*} 0 & < \int_1^{k^2 } {\exp \left( {x/k + \sqrt x \log (N/k)} \right)dx} \\ & = 2k^2 \int_{1/k}^1 {\exp \left( {k\left[ {t^2 + tm} \right]} \right)tdt} \\ & \le 2k^2 \int_0^1 {\exp \left( {k\left[ {t^2 + tm} \right]} \right)tdt} \\ & = \frac{{\sqrt \pi }}{2}\left| m \right|e^{ - km^2 /4} k^{3/2} \left( {\operatorname{erfi}\left( {\frac{{\left| m \right|}}{2}\sqrt k } \right) - \operatorname{erfi}\left( {\frac{{\left| m \right| - 2}}{2}\sqrt k } \right)} \right) - k + ke^{ - k(\left| m \right| - 1)} \\ & < \frac{{\sqrt \pi }}{2}\left| m \right|e^{ - km^2 /4} k^{3/2} \operatorname{erfi}\left( {\sqrt k \frac{{\left| m \right|}}{2}} \right) - k + ke^{ - k(\left| m \right| - 1)} \\ & \sim \frac{2}{{m^2 }} + ke^{ - k(\left| m \right| - 1)} \sim \frac{2}{{m^2 }} \end{align*} for large $k$. Here I used the following two-term asymptotics of the imaginary error function: $$ \operatorname{erfi}(z) \sim \frac{{e^{z^2 } }}{{\sqrt \pi }}\left( {\frac{1}{z} + \frac{1}{{2z^3 }}} \right),\quad z\to +\infty. $$ For small values of $k$ the integral is clearly bounded and for large values of $k$ it even tends to zero.

$\endgroup$
1
$\begingroup$

From an algebraic point of view.

We have the antiderivative $$I=\int e^{a \sqrt{x}+\frac{x}{k}}\,dx$$ $$I=k e^{a+\frac{1}{k}} \left(a \sqrt{k} F\left(\frac{a k+2}{2 \sqrt{k}}\right)-1\right)-k e^{(a+1) k} \left(a \sqrt{k} F\left(\frac{(a+2) \sqrt{k}}{2} \right)-1\right)$$ where appears Dawson integral function.

Then, the analytical expression of $f(n,k)$is

$$\color{blue}{f(n,k)=e^{\frac{1}{k}} n \left(\sqrt{k} \log \left(\frac{n}{k}\right) F\left(\frac{k \log \left(\frac{n}{k}\right)+2}{2 \sqrt{k}}\right)-1\right)-}$$ $$\color{blue}{e^k k \left(\frac{n}{k}\right)^k \left(\sqrt{k} \log \left(\frac{n}{k}\right) F\left(\frac{1}{2} \sqrt{k} \left(\log \left(\frac{n}{k}\right)+2\right)\right)-1\right)}$$ Considering $k$ as a continuous variable, the maximum of $f(n,k)$ is attained at $k\sim n+\frac32$ which makes that we can compute exactly the maximum value of the function. The only problem is that the formulae are quite nasty.

What is pleasant is that $$f(n,n)=\left(e^n-e^{\frac{1}{n}}\right) n$$ and, numerically, $$\frac{f\left(n,n+\frac{3}{2}\right)}{f(n,n)} \sim 1+\frac 1{n}$$

If $k$ is an integer, it has been verified that the maximum value of the function happens for $k=n+1$ as soon as $n \geq 3$.

So, the maximum value of the function is $f(n,n+1)$ and a rather good approximation is, for $n \geq 10$, $$f(n,n+1)\sim e^{n+1}\, n^{n+1} \,(n+1)^{-n} $$

$\endgroup$

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.