Confused at a fairly simple step in an improper integral

glmrkl
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Homework Statement



http://puu.sh/fYQQj/12819720c6.png
My question is in the attempt at the solution (Number 3)

2. Homework Equations

The Attempt at a Solution


I know how to get to lim t→∞ 1/(1-p) * (t^(1-p) - 1^(1-p)), I'm not sure what to do to get the 1 instead of 1^(1-p) in the above image
 
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glmrkl said:
I'm not sure what to do to get the 1 instead of 1^(1-p) in the above image
I'm not exactly sure which part you're referring to. But 1 to any power (including 1-p) is just 1.
 
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I was referring to the part highlighted in red which is the simplified form after you solve the integral for x=t and x=1. I was wondering why they had just put 1 instead of 11-p

Well, this is embarrassing :(... thanks nonetheless!
 
At least it let's the rest of us feel superior! (Until we make a similar careless mistake.)
 
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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...

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