Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Another difficult integral/Also question on checking lower limit.

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\displaystyle{r^{2}\int^{\arcsin\frac{r}{t}}_{0^{+}}\frac{\sqrt{2}\sin(\theta-\frac{\pi}{4})\sin(2\arcsin(\frac{t}{r}\sin\theta))}{1-\cos(2\theta)}d\theta + \int^{\arcsin\frac{r}{t}}_{0^{+}}\frac{\cos(\frac{\pi}{4}-\theta)}{1-\cos(2\theta)}d\theta - \int^{\arcsin\frac{r}{t}}_{0^{+}}\arcsin(\frac{t}{r}\sin\theta)d\theta + \int^{\arcsin\frac{r}{t}}_{0^{+}}d\theta}[/itex]

    [itex] t = \sqrt{2}, r = 1[/itex]

    This is once again beyond my skills for quite some time I imagine, but I am extremely curious to know the answer. The number empire integral calculator gives me that the second term approaches -∞ as theta approaches 0. Does this mean then that with 0+ I wind up with -∞ as the answer still? Or since the change in theta gets smaller too does it depend on the derivative? How do I find if there is a lower limit?
     
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted