A strange integral

Soroban · **Posted:** Sat, 19 Jul 2003 17:07:20 UTC

Hello, everyone!

Try this integral:
$\displaystyle{\int \frac{dx}{1 + e^x}}\right$

So far, I've found FOUR ways to integrate it.
And the four answers all look different.
(Proving them equivalent is the second phase of the problem.)

Give it a try.
I dare you to come up with a fifth method!

radagast · **Posted:** Sat, 19 Jul 2003 19:54:11 UTC

Heres a method i got but considering you found FOUR and this one popped into my head first thing...you probably already found it:
(didnt learn latex code yet so bear with me..)

First notice:
d[ln(1+e^x)] / dx = e^x / (1 + e^x)
which can be conveniently broken into:

d[ln(1+e^x)] / dx = 1 - 1 / (1 + e^x)
Integrating both sides:

ln(1 + e^x) = int(1 - 1 / (1 + e^x)) dx
= int(dx) - int(dx / (1 + e^x))
= x - int(dx / (1 + e^x))
Rearranging:
int(dx / (1 + e^x)) = x - ln(1 + e^x) + C cant forget that constant

Soroban · **Posted:** Sat, 19 Jul 2003 20:53:06 UTC

Hello, radagast!

Wow! ----- Method number FIVE!

I love the deviousness of your approach: "We note that ..."
... and the fact that you never integrated it.

I thought my Method #4 was quite clever and bizarre.
But you win -- hands down!

Thank you!

Soroban · **Posted:** Sun, 20 Jul 2003 14:00:33 UTC

I see that there many read this thread, but few reply.

Assuming that some are still working on a response
(and others couldn't possibly care less),
I'll post my four solutions one at a time.

If you're still trying, STOP!

. . . don't read any further!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$Let\ I = \displaystyle{\int\frac{dx}{1 + e^x}}$

Method #1: "Long Division"

$\displaystyle{\frac{1}{1 + e^x} = \frac{1 + {\bf e^x - e^x}}{1 + e^x} = \frac{1 + e^x}{1 + e^x} - \frac{1}{1 + e^x} = 1 - \frac{e^x}{1 + e^x}}$

Then $I = \displaystyle{\int(1 - \frac{e^x}{1 + e^x})dx} = {\bf x - ln(1 + e^x) + C}$ . . . Solution #1

Soroban · **Posted:** Sun, 20 Jul 2003 14:17:47 UTC

If you don't want any hints, don't read further!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Method #2: $\frac{du}{u}$

Let $\displaystyle{I = \int\frac{dx}{1 + e^x}}$

Multiply numerator and denominator by $e^{-x}$

$\displaystyle{\frac{1}{1 + e^x} = \frac{e^{-x}}{e^{-x}}\cdot \frac{1}{1 + e^{-x}} = \frac{e^{-x}}{e^{-x} + 1} = -\frac{-e^{-x}}{e^{-x} + 1}}$

Then $I = \displaystyle{-\int \frac{-e^{-x}}{e^{-x} + 1}dx} = {\bf -ln(e^{-x} + 1) + C}$ . . . Solution #2

Soroban · **Posted:** Sun, 20 Jul 2003 15:18:11 UTC

If you don't want any hints, don't read further!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Method #3: Substitution

We have: $I = \displaystyle\int\frac{dx}{1 + e^x}}$

Let U = 1 + e^x

Then $e^x = U - 1\ \ \rightarrow\ \ x = ln|U - 1|\ \ \rightarrow\ \ dx = \frac{dU}{U - 1}$

Substituting: $I = \displaystyle{\int\frac{dU}{U(U - 1)}}$

Now, we have two choices for completing the integration . . .

(A) Partial Fractions

$\displaystyle{\int\frac{dU}{U(U - 1)} = \int(\frac{1}{U - 1} - \frac{1}{U})dU} = ln|U - 1| - ln|U| + C$

$= \displaystyle{ln\left|\frac{U - 1}{U}\right| + C = {\bf ln\left(\frac{e^x}{1 + e^x}\right) + C}}$

(B) Complete the Square

$\displaystyle{\int\frac{dU}{U^2 - U} = \int\frac{dU}{(U - \frac{1}{2})^2 - (\frac{1}{2})^2} = ln\left|\frac{(U - \frac{1}{2}) - \frac{1}{2}}{(U - \frac{1}{2}) + \frac{1}{2}}\right| + C}$

$= \displaystyle{ln \left|\frac{U - 1}{U}\right| + C = {\bf ln\left(\frac{e^x}{1 + e^x}\right) + C}}$ . . . Solution #3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note: the substitution u = e^x

leads to the same ntegral.

Soroban · **Posted:** Sun, 20 Jul 2003 16:45:36 UTC

If you don't want any hints, don't read any further!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Method #4: Trig Substitution

We have: $I = \displaystyle{\int\frac{dx}{1 + e^x}}$

Let $e^x = \tan^2\theta$

Then the denominator is: $1 + e^x = 1 + \tan^2\theta = \sec^2\theta$

Also, $e^\frac{x}{2}} = \tan\theta\ \ \rightarrow\ \ \frac{x}{2} = ln(\tan\theta)\ \ \rightarrow\ \ dx = 2\cdot\frac{\sec^2\theta}{\tan\theta}d\theta$

Substituting, we get:

$I = \displaystyle{\int \frac{1}{\sec^2\theta}\cdot 2\cdot \frac{\sec^2\theta}{\tan\theta}d\theta} = 2\int \cot\theta\ d\theta = 2\cdot ln|\sin\theta| + C$

Since $\tan\theta = e^{\frac{x}{2}}$ , then angle $\theta$ is in a right triangle with "opp" = $e^{\frac{x}{2}}$ , "adj" = 1,
and "hyp" = $\sqrt{1 + e^x}$ . Hence, $\sin\theta = \frac{e^{\frac{x}{2}}}{\sqrt{1 + e^x}}$

Therefore: $I = {\bf \displaystyle{2\cdot ln\left(\frac{e^{\frac{x}{2}}}{\sqrt{1 + e^x}}\right) + C}}$ . . . Solution #4

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I had exhausted my imagination at this point.
(Amazing what the brain can do when it has way too much free time!)

Then Radagast surprised me with a stunning fifth method.

Is there a sixth . . . or more?'

[And don't forget to verify that solutions #1, 2, 3, 4 are equivalent!]

skeeter · **Posted:** Sun, 20 Jul 2003 21:48:02 UTC

let u^2 = e^x
2u*du = e^x*dx
2u*du = u^2*dx
2/u*du = dx

INT[1/(1+e^x)dx] =
INT[1/(1+u^2)*2/u*du]
INT[2/[u(1+u^2)*du] =
INT[2/u*du] - INT[2u/(1+u^2)*du] =
2ln|u| - ln|1+u^2| + C =
ln[u^2/(1+u^2)] + C =
ln[e^x/(1+e^x)] + C

Soroban · **Posted:** Mon, 21 Jul 2003 00:42:33 UTC

Well done, skeeter!
And thank you!

dcl · **Joined:** Fri, 23 May 2003 11:09:48 UTC **Posts:** 206

heheh gw

Soroban, you sound pretty bored man... Maybe you wanna solve a quartic for using the quartic formula Mother Goose posted?

here's one of the top of my head.

2x^4 - 3x^3 + x^2 - 8x - 14

have a shot, showing all working

mwhahahahahar.

j/k........or am I?

Mother Goose · **Posted:** Wed, 23 Jul 2003 02:49:24 UTC

dcl wrote:

heheh gw

Soroban, you sound pretty bored man... Maybe you wanna solve a quartic for using the quartic formula Mother Goose posted?

here's one of the top of my head.

2x^4 - 3x^3 + x^2 - 8x - 14

have a shot, showing all working

mwhahahahahar.

j/k........or am I?

I don't htink he's that bored. But it would be pretty neat to solve a quartic using the formula using a grease pencil on his windows :roll:

His neighbours would think he'd be going nuts

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A strange integral

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