# Find f and 'calculate' the integral?

## Given $f : \left(0 , + \infty\right) \to \mathbb{R}$, differentiable with ${\lim}_{x \to 0} f \left(x\right) = + \infty$ ${e}^{f} \left(x\right) + f ' \left(x\right) + 1 = 0$ , $\textcolor{w h i t e}{a a}$ $\forall x > 0$ Find $f$ and show that ${\int}_{\ln} {2}^{1} \left({e}^{f} \left(x\right) \left(x + 1\right)\right) \mathrm{dx} < \ln \left(e - 1\right)$

May 29, 2018

See below

#### Explanation:

${e}^{f} \left(x\right) + f ' \left(x\right) + 1 = 0$

${e}^{y} + y ' + 1 = 0 , q \quad y = f \left(x\right)$

$y ' = - 1 - {e}^{y}$

$\frac{\mathrm{dy}}{1 + {e}^{y}} = - \mathrm{dx}$

$z = {e}^{y} , q \quad \mathrm{dz} = {e}^{y} \setminus \mathrm{dy} = z \setminus \mathrm{dy}$

$\int \frac{\mathrm{dz}}{z \left(1 + z\right)} = - \int \mathrm{dx}$

$\int \mathrm{dz} \setminus \frac{1}{z} - \frac{1}{1 + z} = - \int \mathrm{dx}$

$\ln \left(\frac{z}{1 + z}\right) = C - x$

${e}^{y} / \left(1 + {e}^{y}\right) = {e}^{C - x}$

Using the IV:

• ${e}^{C - x} = \frac{1}{{e}^{- y} + 1}$

• ${\lim}_{x \to 0} y = + \infty \implies C = 0$

${e}^{y} \left(1 - {e}^{- x}\right) = {e}^{- x}$

${e}^{y} = {e}^{- x} / \left(1 - {e}^{- x}\right) = \frac{1}{{e}^{x} - 1}$

$y = \ln \left(\frac{1}{{e}^{x} - 1}\right)$

The SHOW bit

$I = {\int}_{\ln 2}^{1} {e}^{y} \left(x + 1\right) \setminus \mathrm{dx}$

$= - {\int}_{\ln 2}^{1} \left(1 + x\right) \left(1 + y '\right) \setminus \mathrm{dx}$

$= - {\int}_{\ln 2}^{1} 1 + x \setminus \mathrm{dx} - \textcolor{red}{{\int}_{\ln 2}^{1} y ' \setminus \mathrm{dx}} - {\int}_{\ln 2}^{1} x y ' \setminus \mathrm{dx}$

$\textcolor{red}{{\int}_{\ln 2}^{1} y ' \setminus \mathrm{dx}} = {\left[\ln \left(\frac{1}{{e}^{x} - 1}\right)\right]}_{\ln 2}^{1} = - \ln \left(e - 1\right)$

$\implies I - \ln \left(e - 1\right) = - {\int}_{\ln 2}^{1} 1 + x \setminus \mathrm{dx} - {\int}_{\ln 2}^{1} x y ' \setminus \mathrm{dx}$

• ${\int}_{\ln 2}^{1} 1 + x \setminus \mathrm{dx} > 0$

• ${\int}_{\ln 2}^{1} x y ' \setminus \mathrm{dx} > 0$

$\implies I < \ln \left(e - 1\right)$

May 29, 2018

$f \left(x\right) = c - x - \ln \left(1 - {e}^{c - x}\right)$

I could not yet demonstrate the inequality, but I found a stronger inequality.

#### Explanation:

Let $g \left(x\right) = {e}^{f \left(x\right)}$ so that, using the chain rule:

$g ' \left(x\right) = f ' \left(x\right) {e}^{f \left(x\right)}$

Note now that:

$f \left(x\right) = \ln \left(g \left(x\right)\right)$,

so:

$f ' \left(x\right) = \frac{g ' \left(x\right)}{g \left(x\right)}$

Substituting in the original equation we have:

$g \left(x\right) + \frac{g ' \left(x\right)}{g \left(x\right)} + 1 = 0$

and as by definition $g \left(x\right) > 0$:

$\frac{\mathrm{dg}}{\mathrm{dx}} + {g}^{2} \left(x\right) + g \left(x\right) = 0$

which is separable:

$\frac{\mathrm{dg}}{\mathrm{dx}} = - {g}^{2} - g$

$\frac{\mathrm{dg}}{g \left(g + 1\right)} = - \mathrm{dx}$

$\int \frac{\mathrm{dg}}{g \left(g + 1\right)} = - \int \mathrm{dx}$

Decomposing the first member using partial fractions:

$\frac{1}{g \left(g + 1\right)} = \frac{1}{g} - \frac{1}{g + 1}$

so:

$\int \frac{\mathrm{dg}}{g} - \int \frac{\mathrm{dg}}{g + 1} = - \int \mathrm{dx}$

$\ln g - \ln \left(g + 1\right) = - x + c$

Using the properties of logarithms:

$\ln \left(\frac{g}{g + 1}\right) = - x + c$

$\frac{g}{g + 1} = {e}^{c - x}$

Now solving for $g$:

$g = {e}^{c - x} \left(g + 1\right)$

$g \left(1 - {e}^{c - x}\right) = {e}^{c - x}$

and finally:

$g \left(x\right) = {e}^{c - x} / \left(1 - {e}^{c - x}\right)$

Now:

$f \left(x\right) = \ln \left(g \left(x\right)\right) = \ln \left({e}^{c - x} / \left(1 - {e}^{c - x}\right)\right) = \ln \left({e}^{c - x}\right) - \ln \left(1 - c {e}^{-} x\right)$

$f \left(x\right) = c - x - \ln \left(1 - {e}^{c - x}\right)$

We can determine $c$ from the condition:

${\lim}_{x \to 0} f \left(x\right) = + \infty$

As:

${\lim}_{x \to 0} c - x - \ln \left(1 - {e}^{c - x}\right) = c - \ln \left(1 - {e}^{c}\right)$

which is finite unless $c = 0$.

Then:

$f \left(x\right) = - x - \ln \left(1 - {e}^{-} x\right)$

Consider now the integral:

${\int}_{\ln 2}^{1} {e}^{f \left(x\right)} \left(x + 1\right) \mathrm{dx} = {\int}_{\ln 2}^{1} {e}^{-} \frac{x}{1 - {e}^{-} x} \left(x + 1\right) \mathrm{dx}$

As:

$\frac{d}{\mathrm{dx}} \left({e}^{-} \frac{x}{1 - {e}^{-} x} \left(x + 1\right)\right) = - \frac{x \cdot {e}^{x} + 1}{{e}^{x} - 1} ^ 2$

we can see that in the interval of integration the function is strictly decreasing, so its maximum value $M$ occurs for $x = \ln 2$:

$M = \left({e}^{-} \ln \frac{2}{1 - {e}^{-} \ln 2}\right) \left(\ln 2 + 1\right) = \frac{\frac{1}{2}}{1 - \frac{1}{2}} \left(\ln 2 + 1\right) = \left(\ln 2 + 1\right)$

Then:

${\int}_{\ln 2}^{1} {e}^{f \left(x\right)} \left(x + 1\right) \mathrm{dx} \le M \left(1 - \ln 2\right)$

${\int}_{\ln 2}^{1} {e}^{f \left(x\right)} \left(x + 1\right) \mathrm{dx} \le 1 - {\ln}^{2} 2$

May 29, 2018

Here is another one

#### Explanation:

a)

${e}^{f} \left(x\right) + f ' \left(x\right) + 1 = 0$ <=>^(*e^(-f(x))

$1 + f ' \left(x\right) {e}^{- f \left(x\right)} + {e}^{- f \left(x\right)} = 0$ $\iff$

$- f ' \left(x\right) {e}^{- f \left(x\right)} = 1 + {e}^{- f \left(x\right)}$ $\iff$

$\left({e}^{- f \left(x\right)}\right) ' = 1 + {e}^{- f \left(x\right)}$ $\iff$

$\left(1 + {e}^{- f \left(x\right)}\right) ' = 1 + {e}^{- f \left(x\right)}$${\iff}^{x > 0}$

so there $c$$\in$$\mathbb{R}$,

$1 + {e}^{- f \left(x\right)} = c {e}^{x}$

• ${\lim}_{x \to 0} {e}^{- f \left(x\right)} {=}_{x \to 0 , y \to - \infty}^{- f \left(x\right) = u} {\lim}_{u \to - \infty} {e}^{u} = 0$

and ${\lim}_{x \to 0} \left(- {e}^{- f \left(x\right)} + 1\right) = {\lim}_{x \to 0} c {e}^{x}$ $\iff$

$c = 1$

Therefore,

$1 + {e}^{- f \left(x\right)} = {e}^{x}$ $\iff$

${e}^{- f \left(x\right)} = {e}^{x} - 1$ $\iff$

$- f \left(x\right) = \ln \left({e}^{x} - 1\right)$ $\iff$

$f \left(x\right) = - \ln \left({e}^{x} - 1\right)$ $\textcolor{w h i t e}{a a}$, $x > 0$

b)

${\int}_{\ln} {2}^{1} \left({e}^{f} \left(x\right) \left(x + 1\right)\right) \mathrm{dx} <$$\ln \left(e - 1\right)$

$f \left(x\right) = - \ln \left({e}^{x} - 1\right)$ ,$x > 0$

$f ' \left(x\right) = - {e}^{x} / \left({e}^{x} - 1\right)$

$- f ' \left(x\right) = {e}^{x} / \left({e}^{x} - 1\right) \ge \frac{x + 1}{{e}^{x} - 1}$ without the ''$=$''

• ${\int}_{\ln} {2}^{1} f ' \left(x\right) \mathrm{dx} > {\int}_{\ln} {2}^{1} \frac{x + 1}{{e}^{x} - 1} \mathrm{dx}$ $\iff$

${\int}_{\ln} {2}^{1} \frac{x + 1}{{e}^{x} - 1} \mathrm{dx} <$$- {\left[f \left(x\right)\right]}_{\ln} {2}^{1} = - f \left(1\right) + f \left(0\right) = \ln \left(e - 1\right)$

However we have

${e}^{f} \left(x\right) \left(x + 1\right) = {e}^{- \ln \left({e}^{x} - 1\right)} \left(x + 1\right) = \frac{x + 1}{{e}^{x} - 1}$

and so , ${\int}_{\ln} {2}^{1} \left(x + 1\right) {e}^{f} \left(x\right) \mathrm{dx} <$$\ln \left(e - 1\right)$