5

What is the value of $\lfloor \int_{0}^{2023}\frac{2}{x+e^x}dx \rfloor$?

My attempt:

If $t$ is smaller than $2023$ and large enough, then $\int_{t}^{2023}\frac{2}{x+e^x}dx\approx\int_{t}^{2023}\frac{2}{e^x}dx=2(e^{-t}-e^{-2023})\approx 2e^{-t}$.
I want to find $t$ such that $\lfloor \int_{0}^{2023}\frac{2}{x+e^x}dx \rfloor=\lfloor \int_{0}^{t}\frac{2}{x+e^x}dx \rfloor$ and $\lfloor \int_{0}^{t}\frac{2}{x+e^x}dx \rfloor$ is easier to evaluate.

Please give me a hint.

AllCatsAreGrey
  • 2,388
佐武五郎
  • 1,718

3 Answers3

5

Note that \begin{align}&\int_0^{2023}\frac2{x+e^x}dx\\ \le&\int_0^{2023}\frac2{e^x}dx\\ =&2(1-e^{-2023})\\ <&2\end{align}

Also, numerical integration shows $\int_0^2\frac2{x+e^x}dx\approx1.38>1$. Therefore, $\int_0^{2023}\frac2{x+e^x}dx>1$ so $\lfloor\int_0^{2023}\frac2{x+e^x}dx\rfloor=1$.

Lucenaposition
  • 2,416
5

For $$\int_0^{2023}\frac2{x+e^x}dx<2$$ you can refer @Lucenaposition's answer. In the following we focus on $$\int_0^{2023}\frac2{x+e^x}dx>1.$$ Use $e^x-1\geq x,\forall x\in\mathbb R$, we can get \begin{align} \int_0^{2023}\frac2{x+e^x}dx >&\int_0^{2}\frac2{2e^x-1}dx\\ =&2\left[\ln\left(2\mathrm{e}^{x}-1\right)-x\right]\bigg|_{0}^{2}\\ =&2\left[\ln\left(2\mathrm{e}^{2}-1\right)-2\right]\\ >&1. \end{align}

Add: The last inequality is due to: $$2\left[\ln\left(2\mathrm{e}^{2}-1\right)-2\right]>1\iff 2\mathrm{e}^{2}-1>\mathrm{e}^{5/2}\\ \iff 4\mathrm{e}^4-4\mathrm{e}^2+1>\mathrm{e}^5,$$ and $$3\mathrm{e}^4>\mathrm{e}^5,\quad \mathrm{e}^4-4\mathrm{e}^2+1> \mathrm{e}^4-4\mathrm{e}^2=(\mathrm{e}^2-4)\mathrm{e}^2>0.$$

Another method: Use the Hermite-Hadamard integral inequalities (maybe too big tool for this question)

If $f''(x)>0$ when $a\le x\le b$, prove that $$(b-a)f\left(\frac{a+b}{2}\right)<\int_{a}^{b}f(x)\;{dx} <(b-a)\frac{f(a)+f(b)}{2}.$$

Let $f(x)=\frac{2}{x+e^x},0\leq x\leq2$, we can check that $$f''(x)>0,\quad 0\leq x\leq2.$$ So $$\int_0^{2023}\frac2{x+e^x}dx >\int_0^{2}\frac2{2e^x-1}dx> 2f(1)=\frac{4}{1+e}>1.$$

Lucenaposition
  • 2,416
Riemann
  • 11,801
  • 1
    Riemann, thank you very much for your answer. But I think $\log(2e^2-1)\geq\log(2)+2$ does not hold. – 佐武五郎 Jul 23 '24 at 05:25
  • 1
    @佐武五郎 Yes ,youar right, But $2\left[\ln\left(2\mathrm{e}^{2}-1\right)-2\right]>1$ is right.I will fix it . – Riemann Jul 23 '24 at 05:30
2

By the Lucenaposition's answer, $\int_{0}^{2023}\frac{2}{x+e^x}dx<2$.
So, $\lfloor \int_{0}^{2023}\frac{2}{x+e^x}dx \rfloor=0$ or $\lfloor \int_{0}^{2023}\frac{2}{x+e^x}dx \rfloor=1$.

My Lemma:
Let $f(x):=\frac{e^x}{x}$ for $x>0$.
Then, $\min f=e$.

Proof of my lemma:
$f'(x)=e^x(\frac{x-1}{x^2})$.
$f'(x)>0$ if $x>1$.
$f'(x)<0$ if $0<x<1$.
So, $\min f=f(1)=e$.

Let $a:=\frac{2e}{1+e}$.
Then, $\frac{a}{2-a}=e$.
So, $\frac{e^x}{x}\geq\frac{a}{2-a}$ for $x>0$.
So, $(2-a)e^x\geq ax$ for $x\geq 0$.
So, $2e^x\geq a(x+e^x)$ for $x\geq 0$.
So, $\frac{2}{x+e^x}\geq a\frac{1}{e^x}$ for $x\geq 0$.

My Lemma 2:
$a(1-e^{-2023})\geq 1$.

Proof of my lemma 2:
$e=e^1=1+\frac{1}{1!}+\dots\geq 2$.
So, $e^{2023}=e\cdot e^{2022}\geq 2\cdot e^{2022}\geq e^{2022}+2$.
So, $e\geq 1+2e^{-2022}$.
So, $2e-2e^{-2022}\geq 1+e$.
So, $2e(1-e^{-2023})\geq 1+e$.
So, $\frac{2e}{1+e}(1-e^{-2023})=a(1-e^{-2023})\geq 1$.

$\int_{0}^{2023} a\frac{1}{e^x}dx=a(1-e^{-2023})\geq 1$.

And $\int_{0}^{2023}\frac{2}{x+e^x}dx\geq\int_{0}^{2023} a\frac{1}{e^x}dx=a(1-e^{-2023})$.

So, $\lfloor \int_{0}^{2023}\frac{2}{x+e^x}dx \rfloor=1$.

佐武五郎
  • 1,718