Tuesday, 17 May 2016

real analysis - Computing limlimitsntoinftyintfracpi20frac(sin(x))n1sin(x),mathrmdx




I would like to compute the following limit, limnπ20(sin(x))n1sin(x)dx.




I am looking for a high school answer.



I tried writing \lim_{n \to \infty} \int_0^{\frac{\pi}{2}}{\frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x = \lim_{n \to \infty} \lim_{ε \to \frac{\pi}{2}}\int_0^ε{\frac{(\sin(x))^n}{1-\sin(x)}}\,\d x},




but it doesn't help me, since 1 - \sin(x) \leq 1, \forall x \in \left[0, \dfrac{\pi}{2}\right].


Answer




Your integral does event convergence, for each n we have \int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}=\infty




In fact Since see here \frac2πx≤\sin x≤x,~~~~~~\forall x \in \left[0, \displaystyle \frac{\pi}{2}\right] we have



\frac{(\frac2πx)^n}{1-\frac2πx}≤\frac{(\sin x)^n}{1-\sin x}≤\frac{x^n}{1-x}\implies \int_0^{\fracπ2}\frac{(\frac2πx)^n}{1-\frac2πx}dx≤\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}≤\int_0^{\fracπ2}\frac{x^n}{1-x}dx
then let u= \frac2πx the we get




\infty=\int_0^{1}\frac{x^n}{1-x}dx≤\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}≤\int_0^{1}\frac{x^n}{1-x}dx+\int_1^{\fracπ2}\frac{x^n}{1-x}dx=\infty



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