We have
I = \int_0^1 \ln x \ln (1-x) dx = \lim_{a\rightarrow 1} \int_0^a \ln x \ln (1-x) dx.
Integrating by part
I= \lim_{a \rightarrow 1}( - \ln (1-a) + \int_0^a \frac{x \ln x}{1-x} - \int_0^a \frac{x}{1-x}).
By performing such a limit we have
I= 2 + \int_0^1 \frac{\ln x}{1-x} dx...
Since e^{iz^2} is an analytic function on \mathbb{C} we have:
\int_{\Gamma_R} e^{iz^2} d z = 0 where \Gamma_R is the "pizza-slice" countour given in the link.
Now
\int_{\Gamma_R} e^{iz^2} d z = \int_0^R e^{i t^2} d t + \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i \theta} d...
Let \epsilon be a small positive real number. We have to prove that there exists a correspondent \delta such as
| e^x - e | < \epsilon for all 1-\delta < x < 1+\delta and x \neq 1.
We have
| e^x - e | < \epsilon \Rightarrow -\epsilon < e^x - e < \epsilon \Rightarrow e -...
The equation you wrote doesn't describe a hyperbola but two straight lines intersecting in the point (-1, 0). In fact you can write
(x+1)^2- 4 y^2 = 0 \Rightarrow (x +1 - 2 y) (x +1 + 2 y) =0
which implies
x+1 -2y=0
or
x + 1 + 2 y =0
(the stright line equations).
If the equation was...
We don't get that! :cry:
I think you have some problems with the fundamental theorems of calculus or with logarithmic properties.
\ln r - \ln r_0 = \ln \frac{r}{r_0}
\ln \omega - \ln \omega_0 = \ln \frac{\omega}{\omega_0}
The integrand function on the left hand side of your equation is wrong ...
From the second equation in post n. 8 you get:
2 \int_{r_0}^r \frac{d \rho}{\rho} = - \int_{\omega_0}^{\omega} \frac{d \omega}{\omega} .