I think it should be zero... the only thing that makes the integral with a Dirac Delta produce something different of zero is that it assumes infinite value at a certain point. For any finite value, which is the case for any function by definition (the Dirac Delta is NOT a function by...
(a+ib) and (c+id) are complex numbers, and, in principle, the trigonometric identities are only proved for real numbers... so doing as you said, you're using complex trigonometric identities to prove complex trigonometric identities.
I think the usual ways to define trigonometric functions in...
1/z is analytic inside the region determined by the curves A and B and on the curves... so, \displaystyle\int_B f(z) dz - \displaystyle\int_A f(z) dz = 0 (the minus sign is to correct orientation of the curves). So, if you found B, you have A.
Joining C with B (or A, it doesn't really...
The problem is: \int \dfrac{1}{\sqrt{x} - \sqrt[3]{x}} dx
Here, http://www.wolframalpha.com/input/?i=integrate+1%2F((x^(1%2F2)-x^(1%2F3))
Click on show steps and that's it.
See LaTeX for formatting your equations here.
s= \dfrac{1}{\displaystyle\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx}
I couldn't solve it. Neither did "Derive 6".
Regards.
Just trying to visualize the formula.
s= \displaystyle\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx )^-1
What about the last missing ")^-1"?