Exercise 8.31

Question

Exercise 31: 
    In the proof of Theorem 7.26 it was shown that $$\int_{-1}^1(1-x^2)^n\,dx\ge\frac{4}{3\sqrt{n}}$$ for $n=1,2,3,\ldots.$ Use Theorem 8.20 and Exercise 30 to show the more precise result
    $$\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^1(1-x^2)^n\,dx=\sqrt{\pi}.$$

analambanomenos · Accepted Answer

\begin{align*}
        \lim_{nightarrow\infty}\sqrt{n}\int_{-1}^1(1-x^2)^n\,dx &= \lim_{nightarrow\infty}2\sqrt{n}\int_0^1(1-x^2)^n\,dx & \hbox{by symmetry} \
        &= \lim_{nightarrow\infty}\sqrt{n}\int_0^1t^{-1/2}(1-t)^n\,dt & \hbox{substituting $t$ for $x^2$} \
        &= \lim_{nightarrow\infty}\frac{\sqrt{n}\,\Gamma\big(\frac{1}{2}\big)\Gamma(n+1)}{\Gamma\big(n+\frac{3}{2}\big)} & \hbox{Theorem 8.20} \
        &= \Gamma\big(	extstyle\frac{1}{2}\big)\displaystyle\lim_{nightarrow\infty}\frac{n^{3/2}\Gamma(n)}{\Gamma\big(n+\frac{3}{2}\big)} & \hbox{Theorem 8.18(a)} \
        &= \Gamma\big(	extstyle\frac{1}{2}\big) & \hbox{Exercise 30} \
        &= \sqrt{\pi} & \hbox{8.21 (99)}
    \end{align*}

Exercise 8.31

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Exercise 8.31

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