What Happens When Evaluating Improper Integrals with Limit?

JIHUN
Messages
1
Reaction score
0

Homework Statement


Evaluating the following formula:
upload_2015-10-23_22-41-31.png

The Attempt at a Solution


Since the integral part is unknown, dividing the case into two: converging and diverging
If converging: the overall value will always be 0
If diverging: ...?
 
Physics news on Phys.org
As you say, if the integral converges for ##x \to 0^+##, the result is always zero. However, even if the integral diverges, it is still possible that one can get a finite value. This depends on the speed of divergence. Try it out with ##f(t) = 1## and ##f(t) = 1/t##. What happens for ##f(t) = 1/t^\alpha, \alpha > 1##?
 
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