覚書

最近見つけたイカツイ積分の解き方

詳細な説明は俺は苦手なので明日辺りには こいつ から記事が上がってることでしょう(圧)
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仕事早くて助かるなぁ、その日中に上がったよ
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\begin{align*} I_n = &\int_{0}^{\pi}\frac{\sin^{2n}x}{(1-2r\cos x+r^{2})^{n+1}}dx \\ I_{n+2} = &\int_{0}^{\pi}\frac{\sin^{2(n+1)}x}{(1-2r\cos x+r^{2})^{n+2}} \cdot \frac{\sin^{2}x}{1-2r\cos x+r^{2}}dx \\ = &-\frac{1}{2r(n+2)}\int_{0}^{\pi}\frac{-2r(n+2)\sin x}{(1-2r\cos x+r^{2})^{n+3}} \cdot \sin^{2(n+1)+1}x \ dx \\ = &-\frac{1}{2r(n+2)}\left[\frac{1}{(1-2r\cos x+r^{2})^{n+2}} \cdot \sin^{2(n+1)+1}x\right]^{\pi}_{0} + \frac{2(n+1)+1}{2r(n+2)}\int_{0}^{\pi}\frac{1}{(1-2r\cos x+r^{2})^{n+2}} \cdot \sin^{2(n+1)}x \cos x \ dx \\ = &\frac{2(n+1)+1}{2r(n+2)}\int_{0}^{\pi}\frac{1}{(1-2r\cos x+r^{2})^{n+2}} \cdot \sin^{2(n+1)}x \cos x \ dx \\ = &\frac{2(n+1)+1}{2r(n+2)}\frac{1}{2r(n+1)}\int_{0}^{\pi}\frac{1}{(1-2r\cos x+r^{2})^{n+1}} \cdot \sin^{2n}x ((2n+1)\cos^2 x - \sin^2 x) \ dx \\ = &\frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}I_n - \frac{2(n+1)+1}{2r(n+2)}\frac{2(n+1)}{2r(n+1)}\int_{0}^{\pi}\frac{1}{(1-2r\cos x+r^{2})^{n+1}} \cdot \sin^{2(n+1)}x \ dx \\ = &\frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}I_n - \frac{2(n+1)+1}{2r(n+2)}\frac{1}{r}(r^2+1)I_{n+1} + \frac{2(n+1)+1}{2r(n+1)}\frac{1}{r} 2r\int_{0}^{\pi}\frac{1}{(1-2r\cos x+r^{2})^{n+2}} \cdot \sin^{2(n+1)}x \cos x \ dx \\ = &\frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}I_n - \frac{2(n+1)+1}{2r(n+2)}\frac{1}{r}(r^2+1)I_{n+1} + 2I_{n+2} \end{align*}
と
\begin{align*} I_{1}=&\int^{\pi}_{0} \frac{\sin^2 t}{(r^2-2r\cos t +1)^2} dt \\ =&\frac{1}{2r}\int^{\pi}_{0} \frac{\cos t}{r^2-2r\cos t +1} dt \\ =&\frac{1}{4r^2}\int^{\pi}_{0} \left(\frac{r^2+1}{r^2-2r\cos t +1}-1\right) dt \\ 4r^2I_{1}=&(r^2+1)I_{0} - \pi \\ \\ I_{0} =& \int^{\pi}_{0} \frac{1}{r^2-2r\cos t +1} dt \\ =& \int^{\infty}_{0} \frac{1}{r^2+1-2r\frac{1-t^2}{1+t^2}} \frac{2}{1+t^2} dt \\ =& 2\int^{\infty}_{0} \frac{1}{(1-r)^2 t^2 + (1+r)^2} dt \\ =& 2\frac{|1+r|}{|1-r|} \frac{1}{|1+r|^2}\int^{\infty}_{0} \frac{1}{t^2 + 1} dt \\ =& \frac{\pi}{|1-r^2|} = \frac{\pi}{1-r^2}[|r|<1] \quad \mathrm{or} \quad \frac{\pi}{r^2-1}[|r|>1] \\ I_{1} =&\frac{1}{4r^2}\left(\frac{1+r^2}{\pm(1-r^2)}\pi - \pi\right) \\ =&\frac{1}{2}\frac{\pi}{1-r^2} \quad \mathrm{or} \quad \frac{1}{2r^2}\frac{\pi}{r^2-1} \end{align*}

まで解いた。

以下奴の処理
\begin{align*} I_{n+2}&=\frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}I_n - \frac{2(n+1)+1}{2r(n+2)}\frac{1}{r}(r^2+1)I_{n+1} + 2I_{n+2} \\ r^2 I_{n+2}&=\frac{2(n+1)+1}{2(n+2)}(r^2+1)I_{n+1} - \frac{2(n+1)+1}{2(n+2)}\frac{2n+1}{2(n+1)}I_n \\ r^2 \left( I_{n+2} -\frac{2(n+1)+1}{2(n+2)}I_{n+1}\right)&=\frac{2(n+1)+1}{2(n+2)}\left(I_{n+1} - \frac{2n+1}{2(n+1)}I_n\right) \\ \left( I_{n+2} -\frac{2(n+1)+1}{2(n+2)}I_{n+1}\right)&=\frac{(2n+3)(2n+4)}{(n+2)^2 4 r^2}\left(I_{n+1} - \frac{2n+1}{2(n+1)}I_n\right) \\ \left( I_{n+1} -\frac{2n+1}{2(n+1)}I_{n}\right)&=\frac{1}{(2r)^{2n}}\frac{(2n+1)!}{n!(n+1)! }\left(I_{1} - \frac{1}{2}I_{0}\right) \\ \\ |r|<1\rm{の場合} \\ I_{n+1} -\frac{2n+1}{2(n+1)}I_{n}&=0 \\ I_{n} &= \frac{(2n)!}{2^{2n}(n!)^2}I_{0}=\frac{\pi}{2^{2n}(1-r^2)}\binom{2n}{n} \\ \\ |r|>1\rm{の場合} \\ \\ I_{n+1} -\frac{2n+1}{2(n+1)}I_{n}&=\frac{1}{2}\frac{\pi}{r^2-1}\frac{1}{(2r)^{2n}}\frac{(2n+1)!}{n!(n+1)! }\left(\frac{1}{r^2}-1\right) \\ &=\frac{2\pi}{(2r)^{2n+2}}\frac{(2n+1)!}{n!(n+1)! } \\ I_{n+1} -\frac{(2n+1)(2n+2)}{4(n+1)^2}I_{n}&=\frac{\pi}{(2r)^{2n+2}}\frac{(2(n+1))!}{((n+1)!)^2 } \\ I_{n+1} -\frac{4^{-(n+1)}\binom{2(n+1)}{n+1}}{4^{-n}\binom{2n}{n}}I_{n}&=\frac{\pi}{r^{2n+2}}4^{-(n+1)}\binom{2(n+1)}{n+1} \\ \frac{1}{4^{-(n+1)}\binom{2(n+1)}{n+1}}I_{n+1} -\frac{1}{4^{-n}\binom{2n}{n}}I_{n}&=\frac{\pi}{r^{2n+2}} \\ \frac{1}{4^{-n}\binom{2n}{n}}I_{n}=I_{0} + \pi\sum^{n}_{k=1}\frac{1}{r^{2k}} \\ \frac{1}{4^{-n}\binom{2n}{n}}I_{n}=\frac{\pi}{r^2-1} + \pi\frac{1-\frac{1}{r^{2n}}}{1-r^2} \\ I_{n}=\frac{\pi}{(2r)^{2n}(r^2-1)}\binom{2n}{n} \end{align*}

結論として

$$\int_{0}^{\pi}\frac{\sin^{2n}x}{(1-2r\cos x+r^{2})^{n+1}}dx=\left\{\begin{matrix}\displaystyle \frac{\pi}{2^{2n}(1-r^2)}\binom{2n}{n} & |r|<1 \\ \displaystyle \frac{\pi}{(2r)^{2n}(r^2-1)}\binom{2n}{n} & |r|>1 \\ \rm{undefined} & r=\pm 1\end{matrix}\right.$$

追記
\begin{align*} I_{n}=&\int^{\pi}_{0} \frac{\sin^{2n+1} t}{(r^2-2r\cos t +1)^{n+1}}\cos^4\frac{t}{2} dt \\ =&\int^{\pi}_{0} \frac{\sin t}{(r^2-2r\cos t +1)^{n+1}} \cdot \sin^{2n}t\cos^4\frac{t}{2} dt \\ =&\frac{1}{2rn}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n}} \cdot \sin^{2n-1}t \left(2n\cos^4\frac{t}{2}\cos t - 2\sin t\cos^3\frac{t}{2}\sin\frac{t}{2}\right) dt \\ =&\frac{1}{rn}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n}} \cdot \sin^{2n-1}t \left((n+1)\cos^4\frac{t}{2} \cos t - \cos^4 \frac{t}{2} \right) dt \\ =&-\frac{1}{rn}I_{n-1} + \frac{n+1}{rn}\int^{\pi}_{0} \frac{\sin t}{(r^2-2r\cos t +1)^{n}} \cdot \sin^{2n-2}t \cos^4\frac{t}{2} \cos t dt \\ =&-\frac{1}{rn}I_{n-1} + \frac{n+1}{2 r^2 n(n-1)}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n-1}} \cdot \sin^{2n-3} t \left(2(n-1)\cos^4\frac{t}{2}\cos^2 t - 2\sin t\cos^3\frac{t}{2}\sin\frac{t}{2}\cos t - \sin^2 t \cos^4\frac{t}{2} \right) dt \\ =&-\frac{1}{rn}I_{n-1} + \frac{n+1}{2 r^2 n(n-1)}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n-1}} \cdot \sin^{2n-3} t \cos^4\frac{t}{2} \left(2n-(2n+1)\sin^2 t - 2\cos t \right) dt \\ =&-\frac{1}{rn}I_{n-1} + \frac{n+1}{r^2 (n-1)}I_{n-2} - \frac{(n+1)(2n+1)}{2 r^2 n(n-1)}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n-1}} \cdot \sin^{2n-1} t \cos^4\frac{t}{2} dt- \frac{n+1}{r^2 n(n-1)}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n-1}} \cdot \sin^{2n-3} t \cos^4\frac{t}{2} \cos t dt \\ =&-\frac{1}{rn}I_{n-1} + \frac{n+1}{r^2 (n-1)}I_{n-2} - \frac{(n+1)(2n+1)}{2 r^2 n(n-1)}(r^2+1)I_{n-1} -\frac{(n+1)(2n+1)}{2 r^2 n(n-1)}2r\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n-2}} \cdot \sin^{2n-1} t \cos^4\frac{t}{2} \cos t dt- \frac{n+1}{r^2 n(n-1)}\int^{\pi}_{0} \frac{1}{(r^2-2r\cos t +1)^{n-1}} \cdot \sin^{2n-3} t \cos^4\frac{t}{2} \cos t dt \\ =&-\frac{1}{rn}I_{n-1} + \frac{n+1}{r^2 (n-1)}I_{n-2} - \frac{(n+1)(2n+1)}{2 r^2 n(n-1)}(r^2+1)I_{n-1} -\frac{(n+1)(2n+1)}{2 r^2 n(n-1)}2r\left(\frac{rn}{n+1} I_{n}+\frac{1}{n+1}I_{n-1}\right) - \frac{n+1}{r^2 n(n-1)}\left(\frac{r(n-1)}{n} I_{n-1}+\frac{1}{n}I_{n-2}\right) \end{align*}
とか応用も効く手法のようだが見ての通り漸化式が気持ちわるいのでこだわりでもなけりゃ この記事 のように近道しましょう。