# Integral calculation help

1. Feb 24, 2006

### 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

2. Feb 24, 2006

### StatusX

But you don't have an e-x in the numerator. Try multiplying the top and bottom by e2x, then using the substitution u=1+ex, 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+ex)<1 for any real x.

3. Feb 24, 2006

### 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

4. Feb 24, 2006

### happyg1

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

5. Feb 24, 2006

### arildno

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

6. Feb 24, 2006

### happyg1

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