x・exp(-α x^2)の広義積分[0,∞]

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

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