antiderivative involving euler's formula

• Dec 1st 2009, 09:50 AM
rainer
antiderivative involving euler's formula
I need help with the following:

$\displaystyle \int \frac{1}{\cos{x}+i\sin{x}} dx$ = ?

Thanks
• Dec 1st 2009, 12:47 PM
Aryth
$\displaystyle \int \frac{1}{cosx + isinx}~dx$

$\displaystyle = \int \frac{1}{e^{ix}}~dx$

$\displaystyle \left(e^{ix} = cosx + isinx\right)$ <--- Euler's Formula

$\displaystyle =\int e^{-ix} ~dx$

$\displaystyle =\int cosx - isinx ~dx$

$\displaystyle \left(e^{-ix} = e^{i(-x)} = cos(-x) + isin(-x) = cosx - isinx\right)$ <--- Euler's Formula for -x

$\displaystyle = sinx + icosx$
• Dec 1st 2009, 01:57 PM
Drexel28
Quote:

Originally Posted by rainer
I need help with the following:

$\displaystyle \int \frac{1}{\cos{x}+i\sin{x}} dx$ = ?

Thanks

You don't exactly need Euler's formula here. Multiply by the algebraic conjugate of denominator and watch the problem melt away.
• Dec 1st 2009, 03:38 PM
rainer
But wait, so you guys are saying that

$\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$ ?

and that, since the antiderivative of e^x is e^x,

$\displaystyle \frac{1}{e^{ix}}= e^{ix}$

?

That would contradict $\displaystyle \ln{x}=\int_{1}^{x}\frac{1}{x}~dx$

wouldn't it?
• Dec 1st 2009, 03:47 PM
Defunkt
Quote:

Originally Posted by rainer
But wait, so you guys are saying that

$\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$ ?

and that, since the antiderivative of e^x is e^x,

$\displaystyle \frac{1}{e^{ix}}= e^{ix}$

?

That would contradict $\displaystyle \ln{x}=\int_{1}^{x}\frac{1}{x}~dx$

wouldn't it?

Nothing in either post implies $\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$. Both ways give the result of

$\displaystyle \int \frac{1}{e^{ix}}~dx = sinx + icosx \neq cosx + isinx = e^{ix}$
• Dec 2nd 2009, 06:40 AM
rainer
Quote:

Originally Posted by Defunkt
Nothing in either post implies $\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$. Both ways give the result of

$\displaystyle \int \frac{1}{e^{ix}}~dx = sinx + icosx \neq cosx + isinx = e^{ix}$

Okay sorry. I spazzed out there for a second. Now I got it.

But while we're on the subject, the wikipedia page on i does say that i= -i.

If that is true, then wouldn't it follow that:

$\displaystyle \frac{1}{e^{ix}}=e^{ix}$ ?

The wikipedia quote Verbatim:
"If all mathematical textbooks and published literature referring to imaginary or complex numbers were rewritten with −i replacing every occurrence of +i (and therefore every occurrence of −i replaced by −(−i) = +i), all facts and theorems would continue to be equivalently valid"
• Dec 2nd 2009, 06:41 AM
Drexel28
Quote:

Originally Posted by rainer
Okay sorry. I spazzed out there for a second. Now I got it.

But while we're on the subject, the wikipedia page on i does say that i= -i.

If that is true, then wouldn't it follow that:

$\displaystyle \frac{1}{e^{ix}}=e^{ix}$ ?

The wikipedia quote Verbatim:
"If all mathematical textbooks and published literature referring to imaginary or complex numbers were rewritten with −i replacing every occurrence of +i (and therefore every occurrence of −i replaced by −(−i) = +i), all facts and theorems would continue to be equivalently valid"

$\displaystyle i=-i=i^3\implies i^2=1$. Sweet!

Can you find the subtle difference in what was meant to be said and what was said?
• Dec 2nd 2009, 06:54 AM
rainer
Say what?
• Dec 2nd 2009, 06:57 AM
Drexel28
Quote:

Originally Posted by rainer
Say what?

Basically the concept is that $\displaystyle i$ arose from solving $\displaystyle x^2+1=0$. But the FTA tells us that there must be two solutions, obviously $\displaystyle x=i,-i$. As a convention then we take $\displaystyle \sqrt{-1}=i$ (sort of). We could have, with equal validity, make the convention that $\displaystyle \sqrt{-1}=-i$ in which case everything would be different. But this is no way means that $\displaystyle i=-i$