Integral

**liyi** · September 21st 2007, 01:42 PM

Compute $\int_{-1}^1\frac{x^2}{1+e^x}\,dx$

I can't get it!

**Krizalid** · September 21st 2007, 01:45 PM

Facts you should know

$\frac1{1+e^x}+\frac1{1+e^{-x}}=1$

Set $u=-x$

**topsquark** · September 21st 2007, 01:57 PM

Originally Posted by Krizalid

Facts you should know

$\frac1{1+e^x}+\frac1{1+e^{-x}}=1$

Set $u=-x$

Clever. No Physicist ever taught me that kind of trick when doing this integral. (It crops up in integrals over fermionic particles in solid state Physics.)

-Dan

**Jhevon** · September 21st 2007, 02:14 PM

Originally Posted by Krizalid

Facts you should know

$\frac1{1+e^x}+\frac1{1+e^{-x}}=1$

Set $u=-x$

...umm...i don't get it. how do we apply his exactly?

**topsquark** · September 21st 2007, 02:18 PM

Originally Posted by liyi

Compute $\int_{-1}^1\frac{x^2}{1+e^x}\,dx$

I can't get it!

Originally Posted by Jhevon

...umm...i don't get it. how do we apply his exactly?

Look at the u = -x case:
$\int_1^{-1}\frac{(-u)^2}{1+e^{-u}}\,(-du)$

$= \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx$

Now add the two integrals. (Hint: Do an integration by parts on the second integral and see what you get.)

-Dan

**topsquark** · September 21st 2007, 02:21 PM

Originally Posted by Krizalid

Facts you should know

$\frac1{1+e^x}+\frac1{1+e^{-x}}=1$

Set $u=-x$

On second thought this trick isn't likely to work well in Physics. The reason this works here is because of the symmetric limits, and the Physics integrals go from 0 to $\infty$ .

It's still a really cool way to do this integral, though.

-Dan

**Jhevon** · September 21st 2007, 02:54 PM

Originally Posted by topsquark

Look at the u = -x case:
$\int_1^{-1}\frac{(-u)^2}{1+e^{-u}}\,(-du)$

$= \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx$

Now add the two integrals. (Hint: Do an integration by parts on the second integral and see what you get.)

-Dan

ok, i'm not getting anywhere with this:

$\int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx = -x^2 \ln \left( e^{-x} + 1 \right) + 2 \int x \ln \left( e^{-x} + 1 \right) ~dx$

i also did by parts on the last integral also, but it never seemed to be leading me anywhere. and even if i could get a nice result, i'm not exactly sure how you expect me to combine the two integrals when they have different variables.

**Krizalid** · September 21st 2007, 05:08 PM

Originally Posted by Jhevon

$\int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx$

This is the key, 'cause we have the same integration limits, so we can turn back the $u$ into $x$

Now

$\int_{-1}^1\frac{x^2}{1+e^x}\,dx+\int_{-1}^1\frac{x^2}{1+e^{-x}}\,dx=\frac13x^3\bigg|_{-1}^{\phantom{-}1}=\frac23$

Divide by two this last result and yields the desired.

**topsquark** · September 21st 2007, 05:46 PM

Originally Posted by Jhevon

ok, i'm not getting anywhere with this:

$\int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx = -x^2 \ln \left( e^{-x} + 1 \right) + 2 \int x \ln \left( e^{-x} + 1 \right) ~dx$

i also did by parts on the last integral also, but it never seemed to be leading me anywhere. and even if i could get a nice result, i'm not exactly sure how you expect me to combine the two integrals when they have different variables.

Hmmm...

I plugged the integral in on my calculator and got that
$\int \frac{x^2}{1 + e^{-x}}dx = \frac{1}{3}x^3 - \int \frac{x^2}{1 + e^x} dx$

At a glance it looked like an integration by parts, but I guess it isn't. (Or, rather, it still may be but I can't figure how to do it.) I'll have to work on the integral a bit, but in any event you should be able to see where to go with it from here.

Edit: Ah wait! I see where Krizalid is going with this. The integrand isn't an even function, but the two are reflections of each other over the y-axis. Since the integration limits are symmetric the two integrals must be equal to each other.

-Dan

**Jhevon** · September 22nd 2007, 03:34 PM

Originally Posted by Krizalid

This is the key, 'cause we have the same integration limits, so we can turn back the $u$ into $x$

Now

$\int_{-1}^1\frac{x^2}{1+e^x}\,dx+\int_{-1}^1\frac{x^2}{1+e^{-x}}\,dx=\frac13x^3\bigg|_{-1}^{\phantom{-}1}=\frac23$

Divide by two this last result and yields the desired.

Yes! that was the key! Got it! now i really appreciate the ingenious of your substitution, i wouldn't thought of that in a million years! how did you "see" that?!

**Krizalid** · September 22nd 2007, 03:36 PM

It is just practice.

Since I like integrals too

**Jhevon** · September 22nd 2007, 03:40 PM

Originally Posted by Krizalid

It is just practice.

no, it's because you are the man!!!!

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