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I couldn't solve this. $$\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx$$

I thought that like the following.

$$\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx=\int ^{1}_{-1}\dfrac {1-\left( 1-x^{2n}\right) }{\sqrt {1-x^{2}}}dx\\=\int ^{1}_{-1}\dfrac {1-\sqrt {\left( 1-x^{2n}\right) ^{2}}}{\sqrt {1-x^{2}}}dx\\=\int ^{1}_{-1}\dfrac {1}{\sqrt {1-x^{2}}}dx-\int ^{1}_{-1}\dfrac {\sqrt {\left( 1-x^{2n}\right) ^{2}}}{\sqrt {1-x^{2}}}dx$$

I don't know what to do next. Maybe the procedure so far is wrong. Please tell me how to solve.

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    $\begingroup$ Why not put $x=\sin t$? $\endgroup$ Jul 13, 2020 at 4:31
  • $\begingroup$ Plugging it into Wolfram gives a rather ugly expression involving the gamma function, so I suspect that @AnginaSeng's substitution won't solve this for general $n$ $\endgroup$
    – boink
    Jul 13, 2020 at 4:36

3 Answers 3

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Using $$I_n=\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}\,dx=2\int_0^{\frac{\pi}{2}}\sin^{2n}(t)\,dt=\sqrt{\pi }\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}=4^{-n} \binom{2 n}{n}\pi$$

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  • $\begingroup$ Thank you very much! $\endgroup$
    – langhtorn
    Jul 13, 2020 at 14:08
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Either set $x= \sin y$ or then we have $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2n} y =2\int_0^{\frac{\pi}{2}}\sin^{2n}y dy\\=\pi{2n\choose n}\frac{1}{4^n} $$ notice that the last integral we have is Walli's Integral.

Without subbing $$\int_{-1}^{1}x^{2n} (1-x^2)^{\frac{-1}{2}} dx=2\int_0^1x^{2n} \left(\sum_{k=0}^{\infty} (-1)^k {-\frac{1}{2}\choose k} x^{2k}\right) dx \\=2\sum_{k=0}^{\infty}(-1)^{k}{-\frac{1}{2}\choose k}\int_0^1x^{2(n+k)}dx=\sum_{k=0}^{\infty}(-1)^k{-\frac{1}{2} \choose k} \frac{2}{2n+2k+1}$$

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  • $\begingroup$ I would appreciate you look what was your second edit of your answer that is to say before I answered. $\endgroup$ Jul 13, 2020 at 8:44
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    $\begingroup$ Thank you very much! $\endgroup$
    – langhtorn
    Jul 13, 2020 at 14:08
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Let $x=\sin\theta\implies dx=\cos\theta \ d\theta$ $$\int ^{1}_{-1}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx=2\int ^{1}_{0}\dfrac {x^{2n}}{\sqrt {1-x^{2}}}dx$$

$$=2\int ^{\pi/2}_{0}\dfrac {\sin^{2n}\theta}{\cos\theta}\cos\theta \ d\theta$$ $$=2\int ^{\pi/2}_{0}\sin^{2n}\theta\ d\theta$$ Using formula $\color{blue}{\int_0^{\pi/2}\sin^m\theta\cos^n\theta\ d\theta=\dfrac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})}} $, $$=2\frac{\Gamma(\frac{2n+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{2n+0+2}{2})}$$ $$=\frac{\Gamma(n+\frac{1}{2})\sqrt{\pi}}{\Gamma(n+1)}$$

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    $\begingroup$ Thank you very much! I understand that. $\endgroup$
    – langhtorn
    Jul 13, 2020 at 14:07

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