Convergence of improper integrals theorems

R_beta.v3
Messages
13
Reaction score
0

Homework Statement


I'm trying to prove these two theorems
a) if ## 0 \leq f(x) \leq g(x) ## for all x ## \geq 0 ## and ## \int_0^\infty g ## converges, then ## \int_0^\infty f ## converges
b) if ## \int_0^\infty |f| ## converges then ## \int_0^\infty f ## converges.

Obviously assuming f is integrable on every interval [0, N], N ## \geq 0##.

Homework Equations




The Attempt at a Solution


For a).
Let F be defined by ##F(x) = \int_0^x f ##. ##F## is bounded above by ## \int_0^\infty g ##. and since, ##0 \leq f(x)## for all x, F is non-decreasing. So ##\displaystyle\lim_{x\rightarrow \infty} F(x)## exists.

For b)
##0 \leq f(x) + |f(x)| \leq 2|f(x)|## for all x. and ##\int_0^\infty 2|f| ## is convergent, so ##\int_0^\infty (f + |f|) ## is convergent by (a).
The existence of ## \displaystyle\lim_{x\rightarrow \infty} \int_0^x |f|## and the existence of ##\displaystyle\lim_{x\rightarrow \infty} \int_0^x (f + |f|) = \displaystyle\lim_{x\rightarrow \infty}\left( \int_0^x f + \int_0^x |f| \right)##. implies the existence of ##\displaystyle\lim_{x\rightarrow \infty} \int_0^x f ##
 
Physics news on Phys.org
R_beta.v3 said:

Homework Statement


I'm trying to prove these two theorems
a) if ## 0 \leq f(x) \leq g(x) ## for all x ## \geq 0 ## and ## \int_0^\infty g ## converges, then ## \int_0^\infty f ## converges
b) if ## \int_0^\infty |f| ## converges then ## \int_0^\infty f ## converges.

Obviously assuming f is integrable on every interval [0, N], N ## \geq 0##.

Homework Equations




The Attempt at a Solution


For a).
Let F be defined by ##F(x) = \int_0^x f ##. ##F## is bounded above by ## \int_0^\infty g ##. and since, ##0 \leq f(x)## for all x, F is non-decreasing. So ##\displaystyle\lim_{x\rightarrow \infty} F(x)## exists.

For b)
##0 \leq f(x) + |f(x)| \leq 2|f(x)|## for all x. and ##\int_0^\infty 2|f| ## is convergent, so ##\int_0^\infty (f + |f|) ## is convergent by (a).
The existence of ## \displaystyle\lim_{x\rightarrow \infty} \int_0^x |f|## and the existence of ##\displaystyle\lim_{x\rightarrow \infty} \int_0^x (f + |f|) = \displaystyle\lim_{x\rightarrow \infty}\left( \int_0^x f + \int_0^x |f| \right)##. implies the existence of ##\displaystyle\lim_{x\rightarrow \infty} \int_0^x f ##

So, what is your question?
 
Sorry. I just want to make sure that my proof is correct? I'm studying by myself.
 
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