x・exp(-α x^2)の広義積分[0,∞]

\(\int_0^{\infty} x e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \sqrt{\frac{t}{\alpha}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty}\left(\frac{t}{\alpha}\right)^{\frac{1}{2}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^{\frac{1}{2}}\sqrt{\alpha}}\int_0^{\infty} \frac{t^{\frac{1}{2}}}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} 1 \cdot e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} t^0 \cdot e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} t^{1-1} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\Gamma\left(1\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha}\;\left(1-1\right)!\;\cdots\;\Gamma(n)=(n-1)!\;(nは自然数) \\&=&\frac{1}{2\alpha}\cdot 1\;\cdots\;0!=1 \\&=&\frac{1}{2\alpha} \end{eqnarray}$$

x^2・exp(-α x^2) の広義積分[0, ∞]

\(\int_0^{\infty} x^2 e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x^2 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \frac{t}{\alpha} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x^2=\frac{t}{\alpha} \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty} \frac{t}{\alpha} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} \frac{t}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t t^{-\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t^{\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t^{\frac{3}{2}-1}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\Gamma\left(\frac{3}{2}\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\Gamma\left(\frac{1}{2}+1\right) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{(2\cdot 1-1)!!}{2^1}\sqrt{\pi}\;\cdots\;\Gamma\left(\frac{1}{2}+n\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}\;(!!は二重階乗) \;(nは自然数) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1!!}{2}\sqrt{\pi} \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1}{2}\sqrt{\pi}\;\cdots\;nの二重階乗は，1 から n まで n と同じ偶奇性を持つものだけを全て掛けた積 \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1\sqrt{\pi}}{2} \\&=&\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}} \end{eqnarray}$$

x^3・exp(-α x^2)の広義積分[0,∞]

\(\int_0^{\infty} x^3 e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x^3 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;x^3=\left(\sqrt{\frac{t}{\alpha}}\right)^3=\left(\frac{t}{\alpha}\right)^{\frac{3}{2}} \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty}\left(\frac{t}{\alpha}\right)^{\frac{3}{2}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^{\frac{3}{2}}\sqrt{\alpha}}\int_0^{\infty} \frac{t^{\frac{3}{2}}}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\int_0^{\infty} t e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\int_0^{\infty} t^{2-1} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\Gamma\left(2\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha^2}\;\left(2-1\right)!\;\cdots\;\Gamma(n)=(n-1)!\;(nは自然数) \\&=&\frac{1}{2\alpha^2}\cdot 1 \\&=&\frac{1}{2\alpha^2} \end{eqnarray}$$