Question

Hey, you guys bored? :D\[\int_0^af\left(x\right)\,dx=\int_{-a}^afrac{f\left(x\right)}{1+e^x}\,dx\]Show that the above is true under the conditions \(f:\mathbb{R}\rightarrow\mathbb{R},f\,\text{is continuous},a\in\mathbb{R},a>0\).

Good luck!

Answer 1

\int_0^af\left(x\right)\,dx=\int_{-a}^a\frac{f\left(x\right)}{1+e^x}\,dx

Answer 2

\[\int_0^af\left(x\right)\,dx=\int_{-a}^a\frac{f\left(x\right)}{1+e^x}\,dx\]

Answer 3

Yes!

Answer 4

interesting-looking :)

Answer 5

Oh, sorry, forgot to specify \(f\,\text{is even}\).

Answer 6

I was about to say there was some even business happening here
that will help I'm sure

Answer 7

1/(1+e^x) is not even

Answer 8

How is \(\displaystyle{\frac{f\left(x\right)}{1+e^x}}\) even?

Answer 9

You're right. It's odd around \(x=0\), \(y=.5\) My bad. Ignore my last comment.

Answer 10

Rewriting the entire question under one post, just cause.

Under the conditions:\[f:\mathbb{R}\rightarrow\mathbb{R},f\,\text{is continuous},f\left(x\right)=f\left(-x\right),a\in\mathbb{R},a>0\]Prove that:\[\int_0^af\left(x\right)\,dx=\int_{-a}^a\frac{f\left(x\right)}{1+e^x}\,dx\]

Answer 11

I am trying a trig sub and interesting things are happening...

Answer 12

hint:

\[\int\limits_{-a}^{a}\frac{f(x)}{1+e^{x}}dx=\int\limits_{-a}^{0}\frac{f(x)}{1+e^{x}}dx+\int\limits_{0}^{a}\frac{f(x)}{1+e^{x}}dx\]

Answer 13

u=-x

Answer 14

\[\int_a^bf\left(x\right)\,dx=\int_a^bf\left(a+b-x\right)\,dx\]This is an important identity for the solution, apparently.

Answer 15

I want to use zarkons tip, I bet it works
I feel I am making progress

Answer 16

I don't suppose anyone knows what this identity is? I'd like to read up its derivation.

Answer 17

\[u=-x\]
\[du=-dx\]

\[\int\limits_{-a}^{0}\frac{f(x)}{1+e^x}dx\]
\[=-\int\limits_{a}^{0}\frac{f(-u)}{1+e^{-u}}du\]
\[=\int\limits_{0}^{a}\frac{f(u)}{1+e^{-u}}du\]
\[=\int\limits_{0}^{a}\frac{f(x)}{1+e^{-x}}dx\]


\[\int\limits_{-a}^{a}\frac{f(x)}{1+e^{x}}dx=\int\limits_{-a}^{0}\frac{f(x)}{1+e^{x}}dx+\int\limits_{0}^{a}\frac{f(x)}{1+e^{x}}dx\]

\[=\int\limits_{0}^{a}\frac{f(x)}{1+e^{-x}}dx+\int\limits_{0}^{a}\frac{f(x)}{1+e^{x}}dx\]

Answer 18

\[=\int\limits_{0}^{a}\frac{f(x)}{1+e^{-x}}dx+\int\limits_{0}^{a}\frac{f(x)}{1+e^{x}}dx\]


\[=\int\limits_{0}^{a}\left[\frac{f(x)}{1+e^{-x}}+\frac{f(x)}{1+e^{x}}\right]dx\]

\[=\int\limits_{0}^{a}\left[\frac{f(x)(1+e^x)+f(x)(1+e^{-x})}{(1+e^{-x})(1+e^x)}\right]dx\]

\[=\int\limits_{0}^{a}\left[\frac{f(x)(2+e^{x}+e^{-x})}{2+e^x+e^{-x}}\right]dx\]

\[=\int\limits_{0}^{a}f(x)dx\]

Answer 19

sorcery I say!
Zarkon rocks :)