If $I=\int_{0}^{\frac 12} \frac{1}{\sqrt{1-x^{2n}}} dx$, then prove that $I<1$ for $n\ge 1$

Question

The maximum value of $\frac{1}{\sqrt{1-x^{2n}}}$ is $\frac{2^n}{\sqrt{2^{2n}-1}}$

So the maximum value for is the integral is $$\frac 12 \frac{2^n}{\sqrt{2^{2n}-1}}$$

Now I don’t know how to find the max value from here. I tried to use $\lim_{n\to \infty} $ but that just gave the answer $\frac 12$

score 2 · Accepted Answer · answered Oct 27 '20 at 15:55

2

For $n\ge1$ and $0\le x\le1$, $x^{2n}\le x^2$ and $\tfrac{1}{\sqrt{1-x^{2n}}}\le\tfrac{1}{\sqrt{1-x^2}}$. So$$I\le\int_0^{1/2}\tfrac{dx}{\sqrt{1-x^2}}=\arcsin\tfrac12=\tfrac{\pi}{6}<1.$$

answered Oct 27 '20 at 15:55

J.G.

115,835

If $I=\int_{0}^{\frac 12} \frac{1}{\sqrt{1-x^{2n}}} dx$, then prove that $I<1$ for $n\ge 1$

1 Answers1