Calculating Integrals with Tricky Substitutions

[happyg1]

hi,
I'm working on thiis:
.25 = \int_{-\infty}^m\frac{e^{x}}{(1+e^{-x})^2}dx
I let u=1+e^{-x} , du=-e{-x}dx
which gives:
\frac{1}{u}=\frac{1}{1+e^{-x}}|_{-\infty}^m
then I solve it out and I get
\frac{1}{1+e^{-m}}-1=.25
I have tried to solve this and I keep getting a negative ln. I moved the 1 to the RHS and then took ln of both sides. I can't get it to work.
help
CC

 

[StatusX]

But you don't have an e^-x in the numerator. Try multiplying the top and bottom by e^2x, then using the substitution u=1+e^x, which will leave you with something like (u-1) to some power divided by u².

By the way, there is no solution to the equation you mention at the end since 1/(1+e^x)<1 for any real x.

 

[happyg1]

ok
Statusx, you are absolutely correct. I have forgotten my - sign. It should read

.25 = \int_{-\infty}^m\frac{e^{-x}}{(1+e^{-x})^2}dx
sorry...
then let

u=1+e^{-x} , du=-e^{-x}dx
which gives
\frac{1}{1+e^{-m}}-1=.25
so then you get
\frac{1}{1+e^{-m}}=1.25
from there I took to ln of both sides and tried to solve for m, but nothing is working out.
CC

 

[happyg1]

then,
1=1.25+1.25e^{-m}
then
-.2=e^{-m}
ln makes no sense
CC

 

[arildno]

It makes no sense because you've been sloppy!
We have:
\int_{-\infty}^{m}\frac{e^{-x}dx}{(1+e^{-x})^{2}}=\frac{1}{1+e^{-x}}\mid_{-\infty}^{m}=\frac{1}{1+e^{-m}}-\frac{1}{1+e^{\infty}}=\frac{1}{1+e^{-m}}

 

[happyg1]

der duh der duh
\frac{1}{e^{-{-\infty}}}=0
I am a sloppy, sloppy student.
I'm all good now.
Thanks