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We invoke a very interesting property of definite integrals given in MSE

If $f(f(x)) = x$ and $f(0) = 1$ then what is the value of $\int_0^1 (x - f(x)) ^{2n} dx$

if $f(f(x))=x, f(0)=1$, then $\int_{0}^1 (x-f(x))^{2n}dx=\frac{1}{2n+1}.$

Here, we choose a rare self inverse function: $f:[0,1] \to [0,1], f(x)=(1-x^k)^{1/k},k>0$

to get a very interesting result as $$\int_{0}^{1} [x-(1-x^k)^{1/k}]^{2n} dx=\frac{1}{2n+1}, k>0, n\in N \dots (*)$$

The question is: How else this integral (*) can be obtained?

Z Ahmed
  • 43,235

1 Answers1

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Let $$I=\int_{0}^{1} [x-(1-x^k)^{1/k}]^{2n} dx, k>0, n\in N$$ Let $x=\sin^{2/k} t \implies dx=(2/k) \sin^{2/k} t \cos t$, then $$I=\int_{0}^{\pi/2} [\sin^{2/k}t-\cos^{2/k}t]^{2n}~(2/k) \sin^{2/k-1}t \cos t ~ dt \dots (1)$$ Changing $t \to \pi/2-t$, we get $$I=\int_{0}^{\pi/2} [\cos^{2/k}t-\sin^{2/k}t]^{2n}~(2/k) \cos^{2/k-1}t \sin t ~ dt \dots (2).$$ By adding last two, we get $$I=\int_{0}^{\pi/2} [\sin^{2/k}t-\cos^{2/k}t]^{2n}~(2/k) [\sin^{2/k-1}t \cos t+\cos^{2/k-1}t \sin t] ~ dt \dots (3)$$ Let $\sin^{2/k} t-\cos^{2/k} t=u$, to get $$2I=\int_{-1}^{1} u^{2n} du \implies I=\frac{1}{2n+1}.$$

Z Ahmed
  • 43,235