Yes, you are correct. The answer to 1^infinity is 1.

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Homework Statement


What's 1^infinity?


Homework Equations


I think the answer is 1 because 1 to the power of anything equals 1, am I right? What's the answer to this?


The Attempt at a Solution

 
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Yes, you are right, although I don't think it's a particularly meaningful question because it treats infinity as a number which is often not really appropriate.
 
You can prove it equals 1 through calculus, but it is actually undefined because infinity isn't a real number.
 
Success said:

Homework Statement


What's 1^infinity?


Homework Equations


I think the answer is 1 because 1 to the power of anything equals 1, am I right? What's the answer to this?


The Attempt at a Solution


That notation is usually shorthand for a limit a^b where a approaches 1 and b approaches infinity. If a is exactly 1, then yes, the limit is 1. Otherwise it's indeterminant (it could be anything) and the limit depends on the exact behavior of a and b.
 
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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