【微分積分】4-6-2 ベータ関数｜問題集
1.次の積分をガンマ関数で表しなさい。
(1)\(\displaystyle \int_0^\infty\frac{x^3}{1+x^5}dx\)
\(\displaystyle =\frac{1}{5}\int_0^\infty\frac{t^{\frac{4}{5}-1}}{1+t}dt\)
\(\displaystyle =\frac{1}{5}B\left(\frac{4}{5},1-\frac{4}{5}\right)\)
\(\displaystyle =\frac{1}{5}\Gamma\left(\frac{4}{5}\right)\Gamma\left(\frac{1}{5}\right)\)
\(\displaystyle =\frac{1}{5}B\left(\frac{4}{5},1-\frac{4}{5}\right)\)
\(\displaystyle =\frac{1}{5}\Gamma\left(\frac{4}{5}\right)\Gamma\left(\frac{1}{5}\right)\)
(2)\(\displaystyle \int_0^1\frac{1}{\sqrt[4]{1-x^4}}dx\)
\(\displaystyle =\frac{1}{4}\int_0^1t^{\frac{1}{4}-1}(1-t)^{-\frac{1}{4}}dt\)
\(\displaystyle =\frac{1}{4}B\left(\frac{1}{4},\frac{3}{4}\right)\)
\(\displaystyle =\frac{1}{4}\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)\)
\(\displaystyle =\frac{1}{4}B\left(\frac{1}{4},\frac{3}{4}\right)\)
\(\displaystyle =\frac{1}{4}\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)\)
(3)\(\displaystyle \int_0^1\frac{x}{\sqrt{1-x^4}}dx\)
\(\displaystyle =\frac{1}{2}\int_0^1(1-t^2)^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{4}B\left(\frac{1}{2},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{4}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}\)
\(\displaystyle =\frac{1}{4}\Gamma\left(\frac{1}{2}\right)^2\)
\(\displaystyle =\frac{1}{4}B\left(\frac{1}{2},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{4}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}\)
\(\displaystyle =\frac{1}{4}\Gamma\left(\frac{1}{2}\right)^2\)
(4)\(\displaystyle \int_0^1\frac{1}{\sqrt{1-x^5}}dx\)
\(\displaystyle =\frac{1}{5}\int_0^1t^{\frac{1}{5}-1}(1-t)^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{5}B\left(\frac{1}{5},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{5}\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(\frac{7}{10})}\)
\(\displaystyle =\frac{\sqrt{\pi}}{5}\frac{\Gamma\left(\frac{1}{5}\right)}{\Gamma\left(\frac{7}{10}\right)}\)
\(\displaystyle =\frac{1}{5}B\left(\frac{1}{5},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{5}\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(\frac{7}{10})}\)
\(\displaystyle =\frac{\sqrt{\pi}}{5}\frac{\Gamma\left(\frac{1}{5}\right)}{\Gamma\left(\frac{7}{10}\right)}\)
(5)\(\displaystyle \int_0^2\frac{x}{\sqrt{2-x}}dx\)
\(\displaystyle =2\sqrt{2}\int_0^1t^{2-1}(1-t)^{-\frac{1}{2}}dt\)
\(\displaystyle =2\sqrt{2}B\left(2,\frac{1}{2}\right)\)
\(\displaystyle =2\sqrt{2}\frac{\Gamma(2)\Gamma\left(\frac{1}{2}\right)}{\Gamma(\frac{5}{2})}\)
\(\displaystyle =2\sqrt{2}B\left(2,\frac{1}{2}\right)\)
\(\displaystyle =2\sqrt{2}\frac{\Gamma(2)\Gamma\left(\frac{1}{2}\right)}{\Gamma(\frac{5}{2})}\)
(6)\(\displaystyle \int_0^\frac{\pi}{2}\sqrt{\cos x}dx\)
\(\displaystyle =\frac{1}{2}\int_0^1t^{-\frac{1}{2}}(1-t)^\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}B\left(\frac{1}{2},\frac{3}{2}\right)\)
\(\displaystyle =\frac{1}{2}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma(2)}\)
\(\displaystyle =\frac{1}{2}B\left(\frac{1}{2},\frac{3}{2}\right)\)
\(\displaystyle =\frac{1}{2}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma(2)}\)
(7)\(\displaystyle \int_0^\frac{\pi}{2}\sin^\alpha xdx\)
\(\displaystyle =\frac{1}{2}\int_0^1t^{\frac{\alpha}{2}-\frac{1}{2}}(1-t)^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{2}B\left(\frac{\alpha+1}{2},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{2}\frac{\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\alpha+2}{2}\right)}\)
\(\displaystyle =\frac{1}{2}B\left(\frac{\alpha+1}{2},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{2}\frac{\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\alpha+2}{2}\right)}\)
(8)\(\displaystyle \int_0^1(1-x^3)^{-\frac{1}{5}}dx\)
\(\displaystyle =\frac{1}{3}\int_0^1t^{\frac{1}{3}-1}(1-t)^{-\frac{1}{5}}dt\)
\(\displaystyle =\frac{1}{3}B\left(\frac{1}{3},\frac{4}{5}\right)\)
\(\displaystyle =\frac{1}{3}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{17}{15}\right)}\)
\(\displaystyle =\frac{1}{3}B\left(\frac{1}{3},\frac{4}{5}\right)\)
\(\displaystyle =\frac{1}{3}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{17}{15}\right)}\)
(9)\(\displaystyle \int_0^\infty\frac{1}{\sqrt{1+x^3}}dx\)
\(\displaystyle =\frac{1}{3}\int_0^\infty t^{-\frac{2}{3}}(1+t)^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{3}B\left(\frac{1}{3},\frac{1}{6}\right)\)
\(\displaystyle =\frac{1}{3}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{1}{2}\right)}\)
\(\displaystyle =\frac{1}{3}B\left(\frac{1}{3},\frac{1}{6}\right)\)
\(\displaystyle =\frac{1}{3}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{1}{2}\right)}\)
(10)\(\displaystyle \int_0^1\frac{x^5}{\sqrt{1-x^4}}dx\)
\(\displaystyle =\frac{1}{4}\int_0^1t^{\frac{1}{2}}(1-t)^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{4}B\left(\frac{3}{2},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{4}\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(2)}\)
\(\displaystyle =\frac{1}{4}B\left(\frac{3}{2},\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{4}\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(2)}\)
次の学習に進もう！