Prove the enclosed area of this function is equal to 1

[Saracen Rue]

Homework Statement

If function ##f## is defined as such that ##f\left(x\right)=x^{\frac{1}{\sum_{n=1}^∞x^n}}##, then prove that the area enclosed between the the derivative function, ##f'(x)##, and the ##x##-axis is equal to ##1## sq unit

Homework Equations

Knowing that the area under a function ##f(x)## between two can be found by using ##\int_b^a\left|f\left(x\right)\right|dx##

The Attempt at a Solution

I'm having quite a lot of trouble with this question. I know that the first step should be to try and find the points at which the function intercepts the x-axis but I'm unsure of how to do this when the function has an infinite sum in it. Any help with this will be greatly appreciated :)

 

[Metmann]

You should first rewrite the exponent. What is ##\sum_{n=1}^{\infty} x^n## for ##|x|>= 1##? For ##|x|<1## you should recognize the geometric series (but be careful: The first term (n=0) is missing here). After rewriting the exponent, you obtain a much simpler x-dependence.

(Result should be: ##f(x) = x^{\frac{1-x}{x}}## for ##|x|<1## and ##f(x) \equiv 1## for ##|x|>=1##. In fact your function is not defined for ##x=-1## but you could define it as ##1## at this point, because it converges from left and right to ##1##)

 

[fresh_42]

Are negative values of ##x## allowed? The function is pretty crazy for them between ##-1 < x < 0##. For real positive values only, the fundamental theorem of calculus should do.

 

[Metmann]

The function is pretty crazy for them between ##-1 < x < 0##.

Indeed, it should even become complex, right?

 

[fresh_42]

Yes.