Integration of Weird Functions - 2

Calculus Level 5

Define f ( x ) f(x) on [ 0 , 1 ] [0,1] by f ( 0 ) = 0 f(0) = 0 and f ( x ) = lim n ( 1 + sin ( π x ) ) n 1 ( 1 + sin ( π x ) ) n + 1 , x ( 0 , 1 ] \large f(x) = \lim_{n\to\infty} \frac{(1+\sin(\frac{\pi}{x}))^{n} -1}{(1+\sin(\frac{\pi}{x}))^{n} +1} , x \in (0,1]

Compute 0 1 f ( x ) d x \large \int_{0}^{1} f(x) \,dx If you think the function is un-integrable then input 1729 1729 as your answer


The answer is -0.386294361.

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1 solution

we have f ( x ) = 0 f(x) = 0 for x = 1 , 1 2 , 1 3 , . . . . . x=1,\frac{1}{2},\frac{1}{3},.....

f ( x ) = 1 f(x) = -1 for 1 2 < x < 1 \frac{1}{2}<x<1

f ( x ) = 1 f(x) = 1 for 1 3 < x < 1 2 \frac{1}{3}<x<\frac{1}{2}

f ( x ) = 1 f(x) = -1 for 1 4 < x < 1 3 \frac{1}{4}<x<\frac{1}{3}

and so on..............

The derived set of points of discontinuity of f is finite. Hence the function is Riemann Integrable .

So we can represent the integral of this piecewise continuous function as an infinite sum .

0 1 f ( x ) d x = 1 ( 1 1 2 ) + 1 ( 1 2 1 3 ) 1 ( 1 3 1 4 ) . . . . . . . . \int_{0}^{1} f(x) dx = -1(1-\frac{1}{2}) + 1(\frac{1}{2} - \frac{1}{3}) -1(\frac{1}{3} - \frac{1}{4}) ........

= ( 1 1 2 + 1 3 + . . . . . . . ) + ( 1 2 1 3 + 1 4 . . . . . . . ) = -(1-\frac{1}{2} +\frac{1}{3}+.......) + (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} -.......)

Using the fact that 1 1 2 + 1 3 1 4 + . . . . . = ln ( 2 ) 1-\frac{1}{2} + \frac{1}{3} -\frac{1}{4} +..... = \ln(2)

We have our answer as 1 2 ln ( 2 ) 1 - 2\ln(2)

PS:- I am not showing the evaluation of the limits as they are very elementary .

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