e^{-px}sin^{2n}x[0,π/2]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n} xdx\\
&=\frac{(2n)!}{p \displaystyle\prod_{k=1}^n (p^2+4k^2)}\left[1-e^{-\frac{pπ}{2}}\left\{1+\displaystyle\sum_{k=1}^n \frac{p^2(p^2+2^2)(p^2+4^2) \cdots \{p^2+(2k-2)^2\}}{(2k)!}\right\}\right]\\
&(2) \displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n+1} xdx\\
&=\frac{(2n+1)!}{\displaystyle\prod_{k=0}^n \{p^2+(2k+1)^2\}}\left[1-pe^{-\frac{pπ}{2}}\left\{1+\displaystyle\sum_{k=1}^n \frac{(p^2+1^2)(p^2+3^2) \cdots \{p^2+(2k-1)^2\}}{(2k+1)!}\right\}\right]\\
\end{alignat}ただし、全て \(p \gt 0, n \in \mathrm{N}\)









<証明>

次の定積分を用います。(詳細はこちらです。)$$\displaystyle\int e^{ax} \sin^n xdx=\frac{e^{ax} \sin^{n-1}x}{n^2+a^2}(a \sin x-n \cos x)+\frac{n(n-1)}{n^2+a^2}\displaystyle\int e^{ax} \sin^{n-2}xdx$$

また、次の式を \(I_n,J_n\) と置くことにします。$$I_n=\frac{-pe^{-\frac{pπ}{2}}}{p^2+n^2},  J_n=\frac{n(n-1)}{p^2+n^2}$$











\((1)\) 式の煩雑さを避けるために \(I_n,J_n\) に置き換えていきます。
\begin{alignat}{2}
&  \displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n} xdx\\
&=\left[\frac{e^{-px} \sin^{2n-1} x}{p^2+(2n)^2}(-p \sin x-2n \cos x)\right]_0^{\frac{π}{2}} +\frac{2n(2n-1)}{p^2+(2n)^2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-2} xdx\\
&=\frac{-pe^{-\frac{pπ}{2}}}{p^2+(2n)^2}+\frac{2n(2n-1)}{p^2+(2n)^2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-2} xdx\\
&=I_{2n}+J_{2n}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-2} xdx\\
&=I_{2n}+J_{2n}\left\{\frac{-pe^{-\frac{pπ}{2}}}{p^2+(2n-2)^2}+\frac{(2n-2)(2n-3)}{p^2+(2n-2)^2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-4} xdx\right\}\\
&=I_{2n}+J_{2n}\left(I_{2n-2}+J_{2n-2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-4} xdx\right)\\
&=I_{2n}+J_{2n}I_{2n-2}+J_{2n}J_{2n-2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-4} xdx\\
&=I_{2n}+J_{2n}I_{2n-2}+J_{2n}J_{2n-2}\left(I_{2n-4}+J_{2n-4}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-6} xdx\right)\\
&\\
&                      \cdots\\
&\\
&=I_{2n}+J_{2n}I_{2n-2}+J_{2n}J_{2n-2}I_{2n-4}+J_{2n}J_{2n-2}J_{2n-4}I_{2n-6}+ \cdots \\
&     \cdots +J_{2n}J_{2n-2}J_{2n-4} \cdots J_4 \left(I_2 +J_2 \displaystyle\int_0^{\frac{π}{2}}e^{-px}dx\right)\\
&=I_{2n}+(J_{2n})I_{2n-2}+(J_{2n}J_{2n-2})I_{2n-4}+(J_{2n}J_{2n-2}J_{2n-4})I_{2n-6}+ \cdots \\
&     \cdots +(J_{2n}J_{2n-2}J_{2n-4} \cdots J_4)I_2 +(J_{2n}J_{2n-2} \cdots J_4J_2) \cdot \frac{1-e^{-\frac{pπ}{2}}}{p}\\
\end{alignat}次の式を \(L\) と置きます。$$L=J_2J_4J_6 \cdots J_{2n-2}J_{2n}=\frac{(2n)!}{(p^2+2^2)(p^2+4^2) \cdots \{p^2+(2n-2)^2\}\{p^2+(2n)^2\}}$$式全体を \(L\) で括ります。$$\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n} xdx=L\left(\frac{1-e^{-\frac{pπ}{2}}}{p}+\frac{I_2}{J_2}+\frac{I_4}{J_2J_4}+\frac{I_6}{J_2J_4J_6}+ \cdots +\frac{I_{2n-2}}{J_2J_4 \cdots J_{2n-2}}+\frac{I_{2n}}{J_2J_4 \cdots J_{2n}}\right)$$
括弧の中をそれぞれ計算していくと
\begin{alignat}{2}
&\frac{I_2}{J_2}=\frac{-pe^{-\frac{pπ}{2}}}{p^2+2^2} \cdot \frac{p^2+2^2}{2 \cdot 1}=\frac{-pe^{-\frac{pπ}{2}}}{2!}\\
&\frac{I_4}{J_2J_4}=\frac{-pe^{-\frac{pπ}{2}}}{p^2+4^2} \cdot \frac{p^2+2^2}{2 \cdot 1} \cdot \frac{p^2+4^2}{4 \cdot 3}=\frac{-pe^{-\frac{pπ}{2}}(p^2+2^2)}{4!}\\
&\frac{I_6}{J_2J_4J_6}=\frac{-pe^{-\frac{pπ}{2}}}{p^2+6^2} \cdot \frac{p^2+2^2}{2 \cdot 1} \cdot \frac{p^2+4^2}{4 \cdot 3}\cdot \frac{p^2+6^2}{6 \cdot 5}=\frac{-pe^{-\frac{pπ}{2}}(p^2+2^2)(p^2+4^2)}{6!}\\
&\\
&                   \cdots\\
&\\
&\frac{I_{2n}}{J_2J_4 \cdots J_{2n}}=\frac{-pe^{-\frac{pπ}{2}}}{p^2+(2n)^2} \cdot \frac{p^2+2^2}{2 \cdot 1} \cdot \frac{p^2+4^2}{4 \cdot 3}\cdot \frac{p^2+6^2}{6 \cdot 5} \cdot \cdots \cdot \frac{p^2+(2n-2)^2}{(2n-2)(2n-3)}\cdot \frac{p^2+(2n)^2}{2n(2n-1)}\\
&          =\frac{-pe^{-\frac{pπ}{2}}(p^2+2^2)(p^2+4^2)\cdots \{p^2+(2n-2)^2\}}{(2n)!}\\
\end{alignat}となるので
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n} xdx=L\left[\frac{1-e^{-\frac{pπ}{2}}}{p}-\frac{pe^{-\frac{pπ}{2}}}{2!}-\frac{pe^{-\frac{pπ}{2}}(p^2+2^2)}{4!}-\frac{pe^{-\frac{pπ}{2}}(p^2+2^2)(p^2+4^2)}{6!}-\cdots -\frac{pe^{-\frac{pπ}{2}}(p^2+2^2)(p^2+4^2)\cdots \{p^2+(2n-2)^2\}}{(2n)!}\right]\\
&                =\frac{L}{p}\left[1-e^{-\frac{pπ}{2}}-\frac{e^{-\frac{pπ}{2}}p^2}{2!}-\frac{e^{-\frac{pπ}{2}}p^2(p^2+2^2)}{4!}-\frac{e^{-\frac{pπ}{2}}p^2(p^2+2^2)(p^2+4^2)}{6!}-\cdots -\frac{e^{-\frac{pπ}{2}}p^2(p^2+2^2)(p^2+4^2)\cdots \{p^2+(2n-2)^2\}}{(2n)!}\right]\\
&                =\frac{L}{p}\left[1-e^{-\frac{pπ}{2}}\left\{1+\frac{p^2}{2!}+\frac{p^2(p^2+2^2)}{4!}+\frac{p^2(p^2+2^2)(p^2+4^2)}{6!}+\cdots +\frac{p^2(p^2+2^2)(p^2+4^2)\cdots \{p^2+(2n-2)^2\}}{(2n)!}\right\}\right]\\
&                =\frac{(2n)!}{p \displaystyle\prod_{k=1}^n (p^2+4k^2)}\left[1-e^{-\frac{pπ}{2}}\left\{1+\displaystyle\sum_{k=1}^n \frac{p^2(p^2+2^2)(p^2+4^2) \cdots \{p^2+(2k-2)^2\}}{(2k)!}\right\}\right]
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n} xdx=\frac{(2n)!}{p \displaystyle\prod_{k=1}^n (p^2+4k^2)}\left[1-e^{-\frac{pπ}{2}}\left\{1+\displaystyle\sum_{k=1}^n \frac{p^2(p^2+2^2)(p^2+4^2) \cdots \{p^2+(2k-2)^2\}}{(2k)!}\right\}\right]$$







\((2)\) \((1)\) と同様に解いていきます。
\begin{alignat}{2}
&  \displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n+1} xdx\\
&=\left[\frac{e^{-px} \sin^{2n} x}{p^2+(2n+1)^2}\{-p \sin x-(2n+1) \cos x\}\right]_0^{\frac{π}{2}} +\frac{(2n+1)(2n)}{p^2+(2n+1)^2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-1} xdx\\
&=\frac{-pe^{-\frac{pπ}{2}}}{p^2+(2n+1)^2}+\frac{(2n+1)(2n)}{p^2+(2n+1)^2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-1} xdx\\
&=I_{2n+1}+J_{2n+1}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-1} xdx\\
&=I_{2n+1}+J_{2n+1}\left\{\frac{-pe^{-\frac{pπ}{2}}}{p^2+(2n-1)^2}+\frac{(2n-1)(2n-2)}{p^2+(2n-1)^2}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-3} xdx\right\}\\
&=I_{2n+1}+J_{2n+1}\left(I_{2n-1}+J_{2n-1}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-3} xdx\right)\\
&=I_{2n+1}+J_{2n+1}I_{2n-1}+J_{2n+1}J_{2n-1}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-3} xdx\\
&=I_{2n+1}+J_{2n+1}I_{2n-1}+J_{2n+1}J_{2n-1}\left(I_{2n-3}+J_{2n-3}\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n-5} xdx\right)\\
&\\
&                      \cdots\\
&\\
&=I_{2n+1}+J_{2n+1}I_{2n-1}+J_{2n+1}J_{2n-1}I_{2n-3}+J_{2n+1}J_{2n-1}J_{2n-3}I_{2n-5}+ \cdots \\
&     \cdots +J_{2n+1}J_{2n-1}J_{2n-3} \cdots J_5 \left(I_3 +J_3 \displaystyle\int_0^{\frac{π}{2}}e^{-px} \sin xdx\right)\\
\end{alignat}一番右の積分は$$\displaystyle\int_0^{\frac{π}{2}}e^{-px} \sin xdx=\left[\frac{e^{-px}}{p^2+1}(-p \sin x – \cos x)\right]_0^{\frac{π}{2}}=\frac{1-pe^{-\frac{pπ}{2}}}{p^2+1}$$となるので
\begin{alignat}{2}
&=I_{2n+1}+(J_{2n+1})I_{2n-1}+(J_{2n+1}J_{2n-1})I_{2n-3}+(J_{2n+1}J_{2n-1}J_{2n-3})I_{2n-5}+ \cdots \\
&     \cdots +(J_{2n+1}J_{2n-1}J_{2n-3} \cdots J_5)I_3 +(J_{2n+1}J_{2n-1} \cdots J_5J_3) \cdot \frac{1-pe^{-\frac{pπ}{2}}}{p^2+1}\\
\end{alignat}次の式を \(M\) と置きます。$$M=\frac{1}{p^2+1} \cdot J_3J_5J_7 \cdots J_{2n-1}J_{2n+1}=\frac{(2n+1)!}{(p^2+1)(p^2+3^2) \cdots \{p^2+(2n-1)^2\}\{p^2+(2n+1)^2\}}$$
式全体を \(M\) で括ります。$$\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n+1} xdx=M\left\{1-pe^{-\frac{pπ}{2}}+\frac{I_3}{J_3} \cdot (p^2+1)+\frac{I_5}{J_3J_5}\cdot (p^2+1)+\frac{I_7}{J_3J_5J_7} \cdot (p^2+1)+ \cdots +\frac{I_{2n-1}}{J_3J_5 \cdots J_{2n-1}}\cdot (p^2+1)+\frac{I_{2n+1}}{J_3J_5 \cdots J_{2n+1}}\cdot (p^2+1)\right\}$$
括弧の中をそれぞれ計算していくと
\begin{alignat}{2}
&\frac{I_3}{J_3}\cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}}{p^2+3^2} \cdot \frac{p^2+3^2}{3 \cdot 2} \cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}(p^2+1)}{3!}\\
&\frac{I_5}{J_3J_5}\cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}}{p^2+5^2} \cdot \frac{p^2+3^2}{3 \cdot 2} \cdot \frac{p^2+5^2}{5 \cdot 4}\cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}(p^2+1)(p^2+3^2)}{5!}\\
&\frac{I_7}{J_3J_5J_7}\cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}}{p^2+7^2} \cdot \frac{p^2+3^2}{3 \cdot 2} \cdot \frac{p^2+5^2}{5 \cdot 4}\cdot \frac{p^2+7^2}{7 \cdot 6} \cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}(p^2+1)(p^2+3^2)(p^2+5^2)}{7!}\\
&\\
&                   \cdots\\
&\\
&\frac{I_{2n+1}}{J_3J_5 \cdots J_{2n+1}}\cdot (p^2+1)=\frac{-pe^{-\frac{pπ}{2}}}{p^2+(2n+1)^2} \cdot \frac{p^2+3^2}{3 \cdot 2} \cdot \frac{p^2+5^2}{5 \cdot 4}\cdot \frac{p^2+7^2}{7 \cdot 6} \cdot \cdots \cdot \frac{p^2+(2n-1)^2}{(2n-1)(2n-2)}\cdot \frac{p^2+(2n+1)^2}{(2n+1)(2n)} \cdot (p^2+1)\\
&                    =\frac{-pe^{-\frac{pπ}{2}}(p^2+1)(p^2+3^2)(p^2+5^2)\cdots \{p^2+(2n-1)^2\}}{(2n+1)!}\\
\end{alignat}となるので
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n+1} xdx=M\left[1-pe^{-\frac{pπ}{2}}-\frac{pe^{-\frac{pπ}{2}}(p^2+1)}{3!}-\frac{pe^{-\frac{pπ}{2}}(p^2+1)(p^2+3^2)}{5!}-\frac{pe^{-\frac{pπ}{2}}(p^2+1)(p^2+3^2)(p^2+5^2)}{7!}-\cdots -\frac{pe^{-\frac{pπ}{2}}(p^2+1)(p^2+3^2)(p^2+5^2)\cdots \{p^2+(2n-1)^2\}}{(2n+1)!}\right]\\
&                 =M\left[1-pe^{-\frac{pπ}{2}}\left\{1+\frac{p^2+1}{3!}+\frac{(p^2+1)(p^2+3^2)}{5!}+\frac{(p^2+1)(p^2+3^2)(p^2+5^2)}{7!}+\cdots +\frac{(p^2+1)(p^2+3^2)(p^2+5^2)\cdots \{p^2+(2n-1)^2\}}{(2n+1)!}\right\}\right]\\
&                 =\frac{(2n+1)!}{\displaystyle\prod_{k=0}^n \{p^2+(2k+1)^2\}}\left[1-pe^{-\frac{pπ}{2}}\left\{1+\displaystyle\sum_{k=1}^n \frac{(p^2+1^2)(p^2+3^2) \cdots \{p^2+(2k-1)^2\}}{(2k+1)!}\right\}\right]
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{2}} e^{-px}\sin^{2n+1} xdx=\frac{(2n+1)!}{\displaystyle\prod_{k=0}^n \{p^2+(2k+1)^2\}}\left[1-pe^{-\frac{pπ}{2}}\left\{1+\displaystyle\sum_{k=1}^n \frac{(p^2+1^2)(p^2+3^2) \cdots \{p^2+(2k-1)^2\}}{(2k+1)!}\right\}\right]$$

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