Understanding the Variables in the Couchy Theorem

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Theorem
transgalactic
Messages
1,386
Reaction score
0
the question and the solution in this link: (you can enlarge by one click on it)
http://img139.imageshack.us/img139/4074/29783975xs2.gif

the key of solving these question is picking the variables
i don't know how they decided to pick those variables?
 
Last edited by a moderator:
Physics news on Phys.org
I can't make heads or tails out of that. You ask "what is the purpose of this process" but I don't see how anyone can tell you that when you haven't said what problem this is supposed to be solving.
 
i need to prove the expression
which is located above the astrix

for that they choose h(x) and y(x)
and make couchy theorem on them
i know that its no completely solved
i know what to do once i build the variables

i can't understand by what logic they chose the h(x)=f(x)/g(x) y(x)=1/g(x)??
 
they are doing some process on the input expression

(which i need to prove) and after that they chose the variables in a certain way
i can't see what the were trying to do in the process
and what lead them to choose this variables??
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top