[Amankris]

ทำตัวนี้ก่อน $\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\,dx =\sqrt{\pi}$

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ให้ $\displaystyle K=\int_{-\infty}^{\infty}e^{-x^2}\,dx $

พิจารณา
$\begin {array}{rcl}
\displaystyle \int_{R^2}e^{-(x^2+y^2)}\,dA &=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\,dx\,dy\\
&=&\left (\int_{-\infty}^{\infty}e^{-x^2}\,dx\right )\left (\int_{-\infty}^{\infty}e^{-y^2}\,dy\right )\\
&=&K^2\\
&&\\
\displaystyle \int_{R^2}e^{-(x^2+y^2)}\,dA &=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\,dx\,dy\\
&=&\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}\,rdr\,d\theta\,\,\,\,\,\,(x=r\cos \theta\,,\,y=r\sin \theta)\\
&=&2\pi\int_{0}^{\infty}\frac{1}{2}e^{-u}\,du\,\,\,\,\,\,(u=r^2)\\
&=&\pi\\
&&\\
K&=&\sqrt{\pi}
\end {array}$

แล้วมาดูคำถาม
$\begin {array}{rcl}
\displaystyle \int_{-\infty}^{\infty}e^{-x^2+x+a}\,dx&=&\int_{-\infty}^{\infty}e^{-(x-\frac{1}{2})^2+b}\,dx\,\,\,\,\,\,(b=a+\frac{1}{4})\\
&=&e^b\int_{-\infty}^{\infty}e^{-v^2}\,dv\,\,\,\,\,\,(v=x-\frac{1}{2})\\
&=&e^b\sqrt{\pi}
\end {array}$