Find f and 'calculate' the integral?

Given f:(0,+oo)->RR, differentiable with

  • lim_(xto0)f(x)=+oo
  • e^f(x)+f'(x)+1=0 , color(white)(aa) AAx>0

Find f and show that int_ln2^1(e^f(x)(x+1))dx<ln(e-1)

3 Answers
Ultrilliam
May 29, 2018

See below

Explanation:

e^f(x)+f'(x)+1=0

e^y+y'+1=0, qquad y = f(x)

y'= - 1 - e^y

(dy)/(1 + e^y) = - dx

z = e^y, qquad dz = e^y \ dy = z \ dy

int (dz)/(z(1 + z)) = - int dx

int dz \ 1/z - 1/(1+z) = - int dx

ln (z/(1+z)) = C - x

e^y/(1+e^y) = e^(C - x)

Using the IV:

  • e^(C - x) = 1/(e^(-y) + 1 )

  • lim_(x to 0) y = +oo implies C = 0

e^y (1 - e^(-x))= e^( - x)

e^y = e^( - x)/(1 - e^(-x)) = 1/(e^x-1)

y = ln( 1/( e^(x)-1))

The SHOW bit

I = int_(ln2)^1 e^y(x+1) \ dx

= - int_(ln2)^1 (1+ x) (1 + y') \ dx

= - int_(ln2)^1 1+ x \ dx -color(red)( int_(ln2)^1 y' \ dx) - int_(ln2)^1 xy' \ dx

color(red)( int_(ln2)^1 y' \ dx) = [ln( 1/( e^(x)-1))]_(ln2)^1 = - ln(e-1)

implies I - ln(e-1) = - int_(ln2)^1 1+ x \ dx - int_(ln2)^1 xy' \ dx

  • int_(ln2)^1 1+ x \ dx gt 0

  • int_(ln2)^1 xy' \ dx gt 0

implies I lt ln(e-1)

Andrea S.
May 29, 2018

f(x) = c -x -ln(1-e^(c-x))

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

Explanation:

Let g(x) = e^(f(x)) so that, using the chain rule:

g'(x) = f'(x) e^(f(x))

Note now that:

f(x) = ln(g(x)),

so:

f'(x) = (g'(x))/(g(x))

Substituting in the original equation we have:

g(x) + (g'(x))/(g(x)) +1 =0

and as by definition g(x) > 0:

(dg)/dx +g^2(x) +g(x) =0

which is separable:

(dg)/dx = -g^2-g

(dg)/(g(g+1)) = -dx

int (dg)/(g(g+1)) = -int dx

Decomposing the first member using partial fractions:

1/(g(g+1)) = 1/g -1/(g+1)

so:

int (dg)/g- int (dg)/(g+1) = -int dx

ln g - ln(g+1) = -x+ c

Using the properties of logarithms:

ln(g/(g+1)) =- x+ c

g/(g+1) = e^(c-x)

Now solving for g:

g = e^(c-x) (g+1)

g(1-e^(c-x)) = e^(c-x)

and finally:

g(x) = e^(c-x)/(1-e^(c-x))

Now:

f(x) = ln(g(x)) = ln(e^(c-x)/(1-e^(c-x))) = ln(e^(c-x)) -ln(1-ce^-x)

f(x) = c -x -ln(1-e^(c-x))

We can determine c from the condition:

lim_(x->0) f(x) = +oo

As:

lim_(x->0) c -x -ln(1-e^(c-x)) = c-ln(1-e^c)

which is finite unless c=0.

Then:

f(x) = -x-ln(1-e^-x)

Consider now the integral:

int_(ln2)^1 e^(f(x))(x+1)dx = int_(ln2)^1 e^-x/(1-e^-x)(x+1)dx

As:

d/dx ( e^-x/(1-e^-x)(x+1) ) = -(x*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=ln2:

M = ( e^-ln2/(1-e^-ln2))(ln2+1) = (1/2)/(1-1/2)(ln2+1) = (ln2+1)

Then:

int_(ln2)^1 e^(f(x))(x+1)dx <= M(1-ln2)

int_(ln2)^1 e^(f(x))(x+1)dx <= 1-ln^2 2

Jim S
May 29, 2018

Here is another one

Explanation:

a)

e^f(x)+f'(x)+1=0 <=>^(*e^(-f(x))

1+f'(x)e^(-f(x))+e^(-f(x))=0 <=>

-f'(x)e^(-f(x))=1+e^(-f(x)) <=>

(e^(-f(x)))'=1+e^(-f(x)) <=>

(1+e^(-f(x)))'=1+e^(-f(x))<=>^(x>0)

so there cinRR,

1+e^(-f(x))=ce^x

  • lim_(xto0)e^(-f(x))=_(xto0,y->-oo)^(-f(x)=u)lim_(uto-oo)e^u=0

and lim_(xto0)(-e^(-f(x))+1)=lim_(xto0)ce^x <=>

c=1

Therefore,

1+e^(-f(x))=e^x <=>

e^(-f(x))=e^x-1 <=>

-f(x)=ln(e^x-1) <=>

f(x)=-ln(e^x-1) color(white)(aa), x>0

b)

int_ln2^1(e^f(x)(x+1))dx<ln(e-1)

f(x)=-ln(e^x-1) ,x>0

f'(x)=-e^x/(e^x-1)

-f'(x)=e^x/(e^x-1)>=(x+1)/(e^x-1) without the ''=''

  • int_ln2^1f'(x)dx>int_ln2^1(x+1)/(e^x-1)dx <=>

int_ln2^1(x+1)/(e^x-1)dx<-[f(x)]_ln2^1=-f(1)+f(0)=ln(e-1)

However we have

e^f(x)(x+1)=e^(-ln(e^x-1))(x+1)=(x+1)/(e^x-1)

and so , int_ln2^1(x+1)e^f(x)dx<ln(e-1)

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