Calculus: Solve indeterminate question up

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Please help ... I forgot how to do these calculus problem...

Does (infinity)^a * e^(-infinity), where a >1 equal to 0?

If so, why? If not, how to calculate it?
 
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Askhwhelp said:
Please help ... I forgot how to do these calculus problem...

Does (infinity)^a * e^(-infinity), where a >1 equal to 0?

If so, why? If not, how to calculate it?

You do these things by finding limits. 'infinity' by itself doesn't mean much. Do you mean limit x->infinity x^a*e^(-x)? Since you have an indeterminant limit exactly how you approach infinity in each instance of 'infinity' is pretty important.
 
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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