Please help me find my mistake in this integration

Dell
Messages
555
Reaction score
0
\intdx/x*(x2-1)0.5 (from 1 to infinity)

i said t=(x2-1)0.5, therefore x2=t2+1
dt=x/(x2-1)0.5

so now i have

\intdx/x*(x2-1)0.5=\intxdx/x2*(x2-1)0.5=\intdt/x2=\intdt/t+1 (now integral from 0 to infinity)

\intdt/t+1 (from 0 to infinity)

=lim arctan(t)^{b}_{0}=pi/4
b-inf


but the correct answer is pi/2, can ANYONE see where i have gone wrong?
 
Physics news on Phys.org
You didn't adjust your limits after the substitution. If t=\sqrt{x^2-1} and x goes from 1 to infinity then t goes from 0 to infinity.

ps. It would be nice if you use brackets properly so we don't have to guess which function is actually being integrated.
 
if you look at the 5th line of text youll see that i did change my limits,
 
I missed that, still are you sure you didn't use the old limits? As they yields pi/4. So what's arctan(0) and what's arctan(x) with x going to infinity? Hint: your mistake has to do with filling in the limits.
 
Last edited:
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