東京工業大学情報理工学院数理・計算科学系 2019年度午前問B

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以下の問いに答えよ．ただし必要ならば

\[ \int_0^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \]

であることは証明なしで使ってよい．

(1) \(s > 0\) に対し，広義積分 \(\int_0^{\infty}e^{-x}x^{s-1}dx\) が収束することを示せ．

(2) \(s > 0\) に対し，ガンマ関数 \(\Gamma(s)\) を \(\Gamma(s) = \int_0^{\infty} e^{-x}x^{s-1}dx\) と定める．このとき \(\Gamma(s+1) = s\Gamma(s)\) を示せ．

(3) ガンマ関数 \(\Gamma(s)\) の \(s = \frac{3}{2}\) での値 \(\Gamma \left( \frac{3}{2} \right)\) を求めよ．

Kai

(1)

According to subadditivity of definite integrals, we split the integral into two as follows

\[ \int_0^{\infty}x^{s-1}e^{-x}dx = \underbrace{\int_0^1 x^{s-1}e^{-x}dx}_{(1)} + \underbrace{\int_1^\infty x^{s-1}e^{-x}dx}_{(2)} \]

For the first integral, we note that since \(s>0\), we can find a segment \([a,A]\) such that \(s \in [a,A]\). Then, for \(s \in [a,A]\) and \(x \in [0, \infty)\) we have

\[ x^{s-1}e^{-x} = \frac{1}{x^{1-s}e^{x}} \leq \frac{1}{x^{1-A}\cdot 1} \]

By comparison test, we observe that the integral (1) converges.

For the second integral we note that

\[ \lim_{x\to\infty}\frac{x^{s-1}e^{-x}}{\frac{1}{x^2}} = \lim_{x\to\infty}x^{s+1}e^{-x} = 0 \]

and again by comparison test, the integral (2) converges.

(2)

\[ \begin{aligned} \Gamma(s+1) &= \int_0^{\infty} e^{-x} x^s dx \\ &= \left[ -e^{-x} x^s \right]_0^{\infty} + \int_0^{\infty} se^{-x}x^{s-1}dx \\ &= s\Gamma(s) \end{aligned} \]

\[ \begin{aligned} \Gamma \left( \frac{3}{2}\right) &= \frac{1}{2} \Gamma \left( \frac{1}{2}\right) \\ &= \frac{1}{2} \int_0^{\infty} e^{-x} x^{-1/2} dx \\ &= \int_0^{\infty} e^{-y^2}dy \\ &= \frac{\sqrt{\pi}}{2} \end{aligned} \]

東京工業大学 情報理工学院 数理・計算科学系 2019年度 午前 問B