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 #c##in##RR#,

#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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