1. ## double improper integral

find $\displaystyle \int_{-\infty}^{\infty} \frac{e^{3 x}}{e^{6 x}+1}dx$

im lost with this one

2. Originally Posted by viet
find $\displaystyle \int_{-\infty}^{\infty} \frac{e^{3 x}}{e^{6 x}+1}dx$

im lost with this one
This is fun.

First find,
$\displaystyle \int \frac{e^{3x}}{e^{6x}+1} dx$
Let $\displaystyle t=e^{3x} \Rightarrow t' = (1/3)e^{3x}$

Thus,
$\displaystyle 3\int \frac{1}{t^2+1} dt = 3\tan^{-1} t +C=3\tan^{-1} e^{3x} + C$

So now,
$\displaystyle \int_{-\infty}^{\infty} \frac{e^{3x}}{e^{6x}+1} = \int_{-\infty}^0 \frac{e^{3x}}{e^{6x}+1} + \int_0^{\infty} \frac{e^{3x}}{e^{6x}+1} dx = 3\tan^{-1} e^x \big|_{-\infty}^0 + 3\tan^{-1} e^x \big|_0^{\infty}$$\displaystyle = \frac{3\pi}{4} - 0 + \frac{3\pi}{4} - 0 =$$\displaystyle \frac{3\pi}{2}$

3. Originally Posted by ThePerfectHacker
First find,
$\displaystyle \int \frac{e^{3x}}{e^{6x}+1} dx$
Let $\displaystyle t=e^{3x} \Rightarrow t' = (1/3)e^{3x}$
Oops, you're not integrating there

4. ## the answer is Pi/6

Originally Posted by ThePerfectHacker
This is fun.

First find,
$\displaystyle \int \frac{e^{3x}}{e^{6x}+1} dx$
Let $\displaystyle t=e^{3x} \Rightarrow t' = (1/3)e^{3x}$

Thus,
$\displaystyle 3\int \frac{1}{t^2+1} dt = 3\tan^{-1} t +C=3\tan^{-1} e^{3x} + C$

So now,
$\displaystyle \int_{-\infty}^{\infty} \frac{e^{3x}}{e^{6x}+1} = \int_{-\infty}^0 \frac{e^{3x}}{e^{6x}+1} + \int_0^{\infty} \frac{e^{3x}}{e^{6x}+1} dx = 3\tan^{-1} e^x \big|_{-\infty}^0 + 3\tan^{-1} e^x \big|_0^{\infty}$$\displaystyle = \frac{3\pi}{4} - 0 + \frac{3\pi}{4} - 0 =$$\displaystyle \frac{3\pi}{2}$
I think the answer is Pi/6.

5. I think the answer is Pi/6.
Yes.