Integration of (e[SUP]-√x[/SUP])/√x

  • Thread starter Thread starter Anne5632
  • Start date Start date
  • Tags Tags
    Integration
Anne5632
Messages
23
Reaction score
2
Homework Statement
Compute the integral with upper bound infinity and lower bound 1
Relevant Equations
Integral of (e[SUP]-√x[/SUP])/√x
(e-√x)/√x (integral from title)

I integrated by substituting and the bounds changed with inf changing to -inf and 1 changing to -1

My final integrated answer is -2lim[e-√x]. What happens to this equation at -inf and -1? As I can't put them into the roots
 
Physics news on Phys.org
I'm not sure I understand the difficulty.
 
Anne5632 said:
Homework Statement:: Compute the integral with upper bound infinity and lower bound 1
Relevant Equations:: Integral of (e-√x)/√x

(e-√x)/√x (integral from title)

I integrated by substituting and the bounds changed with inf changing to -inf and 1 changing to -1

My final integrated answer is -2lim[e-√x]. What happens to this equation at -inf and -1? As I can't put them into the roots
You have the wrong limits. \sqrt{x} is the positive root. Thus \sqrt{1} = 1 and \sqrt{+\infty} = +\infty.
 
pasmith said:
You have the wrong limits. \sqrt{x} is the positive root. Thus \sqrt{1} = 1 and \sqrt{+\infty} = +\infty.
Thank you, I originally let my substitution = -√x
But √x is better
Final answer now is 2/e
 
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