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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$$\begin{array}{rl}{I}_n=& {\int }_0^{\pi }\frac{{\sin }^{2n}x}{(1-2r\cos x+r^2{)}^{n+1}}dx\\ I_{n+2}=& {\int }_0^{\pi }\frac{{\sin }^{2(n+1)}x}{(1-2r\cos x+r^2{)}^{n+2}}\cdot \frac{{\sin }^{2}x}{1-2r\cos x+r^2}dx\\ =& -\frac{1}{2r(n+2)}{\int }_0^{\pi }\frac{-2r(n+2)\sin x}{(1-2r\cos x+r^2{)}^{n+3}}\cdot {\sin }^{2(n+1)+1}xdx\\ =& -\frac{1}{2r(n+2)}{[\frac{1}{(1-2r\cos x+r^2{)}^{n+2}}\cdot {\sin }^{2(n+1)+1}x]}_0^{\pi }+\frac{2(n+1)+1}{2r(n+2)}{\int }_0^{\pi }\frac{1}{(1-2r\cos x+r^2{)}^{n+2}}\cdot {\sin }^{2(n+1)}x\cos xdx\\ =& \frac{2(n+1)+1}{2r(n+2)}{\int }_0^{\pi }\frac{1}{(1-2r\cos x+r^2{)}^{n+2}}\cdot {\sin }^{2(n+1)}x\cos xdx\\ =& \frac{2(n+1)+1}{2r(n+2)}\frac{1}{2r(n+1)}{\int }_0^{\pi }\frac{1}{(1-2r\cos x+r^2{)}^{n+1}}\cdot {\sin }^{2n}x((2n+1){\cos }^{2}x-{\sin }^{2}x)dx\\ =& \frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}{I}_n-\frac{2(n+1)+1}{2r(n+2)}\frac{2(n+1)}{2r(n+1)}{\int }_0^{\pi }\frac{1}{(1-2r\cos x+r^2{)}^{n+1}}\cdot {\sin }^{2(n+1)}xdx\\ =& \frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}{I}_n-\frac{2(n+1)+1}{2r(n+2)}\frac{1}{r}(r^2+1)I_{n+1}+\frac{2(n+1)+1}{2r(n+1)}\frac{1}{r}2r{\int }_0^{\pi }\frac{1}{(1-2r\cos x+r^2{)}^{n+2}}\cdot {\sin }^{2(n+1)}x\cos xdx\\ =& \frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}{I}_n-\frac{2(n+1)+1}{2r(n+2)}\frac{1}{r}(r^2+1)I_{n+1}+2I_{n+2}\end{array}$$
と
$$\begin{array}{rl}{I}_1=& {\int }_0^{\pi }\frac{{\sin }^{2}t}{(r^2-2r\cos t+1{)}^2}dt\\ =& \frac{1}{2r}{\int }_0^{\pi }\frac{\cos t}{r^2-2r\cos t+1}dt\\ =& \frac{1}{4r^2}{\int }_0^{\pi }(\frac{r^2+1}{r^2-2r\cos t+1}-1)dt\\ 4r^2{I}_1=& (r^2+1)I_0-\pi \\ \\ I_0=& {\int }_0^{\pi }\frac{1}{r^2-2r\cos t+1}dt\\ =& {\int }_0^{\mathrm{\infty }}\frac{1}{r^2+1-2r\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt\\ =& 2{\int }_0^{\mathrm{\infty }}\frac{1}{(1-r{)}^2{t}^2+(1+r{)}^2}dt\\ =& 2\frac{|1+r|}{|1-r|}\frac{1}{|1+r{|}^2}{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+1}dt\\ =& \frac{\pi }{|1-r^2|}=\frac{\pi }{1-r^2}[|r|<1]\mathrm{or}\frac{\pi }{r^2-1}[|r|>1]\\ I_1=& \frac{1}{4r^2}(\frac{1+r^2}{\pm (1-r^2)}\pi -\pi )\\ =& \frac{1}{2}\frac{\pi }{1-r^2}\mathrm{or}\frac{1}{2r^2}\frac{\pi }{r^2-1}\end{array}$$

まで解いた。

以下奴の処理
$$\begin{array}{rl}{I}_{n+2}& =\frac{2(n+1)+1}{2r(n+2)}\frac{2n+1}{2r(n+1)}{I}_n-\frac{2(n+1)+1}{2r(n+2)}\frac{1}{r}(r^2+1)I_{n+1}+2I_{n+2}\\ r^2{I}_{n+2}& =\frac{2(n+1)+1}{2(n+2)}(r^2+1)I_{n+1}-\frac{2(n+1)+1}{2(n+2)}\frac{2n+1}{2(n+1)}{I}_n\\ r^2(I_{n+2}-\frac{2(n+1)+1}{2(n+2)}{I}_{n+1})& =\frac{2(n+1)+1}{2(n+2)}(I_{n+1}-\frac{2n+1}{2(n+1)}{I}_n)\\ (I_{n+2}-\frac{2(n+1)+1}{2(n+2)}{I}_{n+1})& =\frac{(2n+3)(2n+4)}{(n+2{)}^{2}4r^2}(I_{n+1}-\frac{2n+1}{2(n+1)}{I}_n)\\ (I_{n+1}-\frac{2n+1}{2(n+1)}{I}_n)& =\frac{1}{(2r{)}^{2n}}\frac{(2n+1)!}{n!(n+1)!}(I_1-\frac{1}{2}{I}_0)\\ \\ |r|<1\mathrm{の}\mathrm{場}\mathrm{合}\\ I_{n+1}-\frac{2n+1}{2(n+1)}{I}_n& =0\\ I_n& =\frac{(2n)!}{2^{2n}(n!{)}^2}{I}_0=\frac{\pi }{2^{2n}(1-r^2)}(\genfrac{}{}{0ex}{}{2n}{n})\\ \\ |r|>1\mathrm{の}\mathrm{場}\mathrm{合}\\ \\ I_{n+1}-\frac{2n+1}{2(n+1)}{I}_n& =\frac{1}{2}\frac{\pi }{r^2-1}\frac{1}{(2r{)}^{2n}}\frac{(2n+1)!}{n!(n+1)!}(\frac{1}{r^2}-1)\\ & =\frac{2\pi }{(2r{)}^{2n+2}}\frac{(2n+1)!}{n!(n+1)!}\\ I_{n+1}-\frac{(2n+1)(2n+2)}{4(n+1{)}^2}{I}_n& =\frac{\pi }{(2r{)}^{2n+2}}\frac{(2(n+1))!}{((n+1)!{)}^2}\\ I_{n+1}-\frac{4^{-(n+1)}(\genfrac{}{}{0ex}{}{2(n+1)}{n+1})}{4^{-n}(\genfrac{}{}{0ex}{}{2n}{n})}{I}_n& =\frac{\pi }{r^{2n+2}}{4}^{-(n+1)}(\genfrac{}{}{0ex}{}{2(n+1)}{n+1})\\ \frac{1}{4^{-(n+1)}(\genfrac{}{}{0ex}{}{2(n+1)}{n+1})}{I}_{n+1}-\frac{1}{4^{-n}(\genfrac{}{}{0ex}{}{2n}{n})}{I}_n& =\frac{\pi }{r^{2n+2}}\\ \frac{1}{4^{-n}(\genfrac{}{}{0ex}{}{2n}{n})}{I}_n=I_0+\pi \sum _{k=1}^n\frac{1}{r^{2k}}\\ \frac{1}{4^{-n}(\genfrac{}{}{0ex}{}{2n}{n})}{I}_n=\frac{\pi }{r^2-1}+\pi \frac{1-\frac{1}{r^{2n}}}{1-r^2}\\ I_n=\frac{\pi }{(2r{)}^{2n}(r^2-1)}(\genfrac{}{}{0ex}{}{2n}{n})\end{array}$$

結論として

$${\int }_0^{\pi }\frac{{\sin }^{2n}x}{(1-2r\cos x+r^2{)}^{n+1}}dx=\{\begin{array}{cc}{\displaystyle \frac{\pi }{2^{2n}(1-r^2)}(\genfrac{}{}{0ex}{}{2n}{n})}& |r|<1\\ {\displaystyle \frac{\pi }{(2r{)}^{2n}(r^2-1)}(\genfrac{}{}{0ex}{}{2n}{n})}& |r|>1\\ \mathrm{u}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d}& r=\pm 1\end{array}$$

追記
$$\begin{array}{rl}{I}_n=& {\int }_0^{\pi }\frac{{\sin }^{2n+1}t}{(r^2-2r\cos t+1{)}^{n+1}}{\cos }^4\frac{t}{2}dt\\ =& {\int }_0^{\pi }\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 }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^n}\cdot {\sin }^{2n-1}t(2n{\cos }^4\frac{t}{2}\cos t-2\sin t{\cos }^3\frac{t}{2}\sin \frac{t}{2})dt\\ =& \frac{1}{rn}{\int }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^n}\cdot {\sin }^{2n-1}t((n+1){\cos }^4\frac{t}{2}\cos t-{\cos }^4\frac{t}{2})dt\\ =& -\frac{1}{rn}{I}_{n-1}+\frac{n+1}{rn}{\int }_0^{\pi }\frac{\sin t}{(r^2-2r\cos t+1{)}^n}\cdot {\sin }^{2n-2}t{\cos }^4\frac{t}{2}\cos tdt\\ =& -\frac{1}{rn}{I}_{n-1}+\frac{n+1}{2r^{2}n(n-1)}{\int }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^{n-1}}\cdot {\sin }^{2n-3}t(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})dt\\ =& -\frac{1}{rn}{I}_{n-1}+\frac{n+1}{2r^{2}n(n-1)}{\int }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^{n-1}}\cdot {\sin }^{2n-3}t{\cos }^4\frac{t}{2}(2n-(2n+1){\sin }^{2}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)}{2r^{2}n(n-1)}{\int }_0^{\pi }\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 }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^{n-1}}\cdot {\sin }^{2n-3}t{\cos }^4\frac{t}{2}\cos tdt\\ =& -\frac{1}{rn}{I}_{n-1}+\frac{n+1}{r^2(n-1)}{I}_{n-2}-\frac{(n+1)(2n+1)}{2r^{2}n(n-1)}(r^2+1)I_{n-1}-\frac{(n+1)(2n+1)}{2r^{2}n(n-1)}2r{\int }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^{n-2}}\cdot {\sin }^{2n-1}t{\cos }^4\frac{t}{2}\cos tdt-\frac{n+1}{r^{2}n(n-1)}{\int }_0^{\pi }\frac{1}{(r^2-2r\cos t+1{)}^{n-1}}\cdot {\sin }^{2n-3}t{\cos }^4\frac{t}{2}\cos tdt\\ =& -\frac{1}{rn}{I}_{n-1}+\frac{n+1}{r^2(n-1)}{I}_{n-2}-\frac{(n+1)(2n+1)}{2r^{2}n(n-1)}(r^2+1)I_{n-1}-\frac{(n+1)(2n+1)}{2r^{2}n(n-1)}2r(\frac{rn}{n+1}{I}_n+\frac{1}{n+1}{I}_{n-1})-\frac{n+1}{r^{2}n(n-1)}(\frac{r(n-1)}{n}{I}_{n-1}+\frac{1}{n}{I}_{n-2})\end{array}$$
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。