積分の問題

こんにちは,itouです.積分の問題がいっぱい載ってるpdfを手に入れたので,やっていきます.

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解答

$$\begin{array}{rl}{\int }_0^1\frac{1}{(1+yx)\sqrt{1-x^2}}dx& ={\int }_0^{\pi /2}\frac{1}{1+y\sin \left(t\right)}dt(x\to \sin \left(t\right))\\ & ={\int }_0^{\pi /2}\frac{(\tan {\left(t/2\right)}^{\prime })}{{\tan }^2(t/2)+2y\tan \left(t/2\right)+1}dt(半角にして分母分子{\cos }^2(t/2)で割る)\\ & =2{\int }_0^1\frac{1}{u^2+2yu+1}du(\tan \left(t/2\right)=u)\\ & =2{\int }_0^1\frac{1}{(u+y{)}^2+(\sqrt{1-y^2}{)}^2}du\\ & =2{\int }_y^{y+1}\frac{1}{t^2+(\sqrt{1-y^2}{)}^2}dt(u+t=t)\\ & =\frac{2}{\sqrt{1-y^2}}(\arctan \left(\frac{y+1}{\sqrt{1-y^2}}\right)-\arctan \left(\frac{y}{\sqrt{1-y^2}}\right))\\ & =\frac{2}{\sqrt{1-y^2}}\arctan \left(\sqrt{\frac{1-y}{1+y}}\right)(\arctan の加法定理)\\ & =\frac{\arccos \left(y\right)}{\sqrt{1-y^2}}(\ast )\end{array}$$

ただし,（＊）では
$$\begin{array}{r}\arccos \left(\alpha \right)=2\arctan \left(\sqrt{\frac{1-\alpha }{1+\alpha }}\right)\end{array}$$
を用いた.

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解答

(1)
$$\begin{array}{r}{I}_{(m,n)}={\int }_0^1{x}^m{\log }^n(x)dx\end{array}$$
とおく.部分積分により漸化式
$$\begin{array}{r}\frac{I_{(m,n)}}{I_{(m,n-1)}}=\frac{-n}{m+1}\end{array}$$
を繰り返し用いて
$$\begin{array}{r}{I}_{(m,n)}={\int }_0^1{x}^m{\log }^n(x)dx=(-1{)}^n\frac{n!}{(m+1{)}^{n+1}}\end{array}$$
を得る.

(2)
$$\begin{array}{r}{\int }_0^a{x}^m{\log }^n(x)dx=a^{m+1}{\int }_0^1{y}^m{\log }^n(ay)dy\end{array}$$

${\log }^n(ay)=(\log \left(a\right)+\log \left(y\right){)}^n$を展開し,(1)を用いることで示される.