【微分積分】4-6-1 ガンマ関数｜問題集
1.次の値を求めなさい。
(1)\(\displaystyle \Gamma(5)\)
\(\displaystyle =4!\)
\(\displaystyle =24\)
\(\displaystyle =24\)
(2)\(\displaystyle \Gamma\left(-\frac{7}{2}\right)\)
\(\displaystyle =-\frac{2}{7}・-\frac{2}{5}・-\frac{2}{3}\Gamma\left(-\frac{1}{2}\right)\)
\(\displaystyle =-\frac{8}{105}・-2\sqrt{\pi}\)
\(\displaystyle =\frac{16}{105}\sqrt{\pi}\)
\(\displaystyle =-\frac{8}{105}・-2\sqrt{\pi}\)
\(\displaystyle =\frac{16}{105}\sqrt{\pi}\)
2.次の値をガンマ関数を使って求めなさい。
(1)\(\displaystyle \int_0^\infty e^{-x^2}dx\)
\(t=x^2\)とおくと、\(\displaystyle dx=\frac{1}{2\sqrt{t}}dt\)
\(\displaystyle =\int_0^\infty e^{-t}\frac{1}{2\sqrt{t}}dt\)
\(\displaystyle =\frac{1}{2}\int_0^\infty t^{-\frac{1}{2}}e^{-t}dt\)
\(\displaystyle =\frac{1}{2}\Gamma\left(\frac{1}{2}\right)\)
\(\displaystyle =\int_0^\infty e^{-t}\frac{1}{2\sqrt{t}}dt\)
\(\displaystyle =\frac{1}{2}\int_0^\infty t^{-\frac{1}{2}}e^{-t}dt\)
\(\displaystyle =\frac{1}{2}\Gamma\left(\frac{1}{2}\right)\)
次の学習に進もう！