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Question Number 188192 by universe last updated on 26/Feb/23

$$$$$$provethat$$$${\int }_0^{\mathrm{\infty }}{e}^{-a^2{x}^2}\cos \left(2bx\right)dx=\frac{\sqrt{\pi }}{2a}{e}^{-b^2/a^2}$$$$$$$$$$

Answered by qaz last updated on 26/Feb/23

$$I\left(b\right)={\int }_0^{\mathrm{\infty }}{e}^{-a^2{x}^2}\cos \left(2bx\right)dx$$$$I{\left(b\right)}^{\prime }=-{\int }_0^{\mathrm{\infty }}2{xe}^{-a^2{x}^2}\sin \left(2bx\right)dx=\frac{1}{a^2}{e}^{-a^2{x}^2}\sin \left(2bx\right){\mid }_0^{\mathrm{\infty }}-\frac{2b}{a^2}{\int }_0^{\mathrm{\infty }}{e}^{-a^2{x}^2}\cos \left(2bx\right)dx$$$$\Rightarrow I{\left(b\right)}^{\prime }=-\frac{2b}{a^2}I\left(b\right)\Rightarrow I\left(b\right)={Ce}^{-\frac{b^2}{a^2}}$$$$C=I\left(0\right)={\int }_0^{\mathrm{\infty }}{e}^{-a^2{x}^2}dx=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2{\left(a^2\right)}^{\frac{1}{2}}}=\frac{\sqrt{\pi }}{2a},a>0$$$$\Rightarrow I\left(b\right)=\frac{\sqrt{\pi }}{2a}{e}^{-\frac{b^2}{a^2}},a>0$$

Commented by universe last updated on 26/Feb/23

$$thankssir$$

Answered by ARUNG_Brandon_MBU last updated on 26/Feb/23

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{e}^{-a^2{x}^2}\cos \left(2bx\right)dx$$$$=\frac{1}{2}Re{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(a^2{x}^2+2ibx)}dx$$$$=\frac{1}{2}Re{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-[a^2{(x+\frac{ib}{a^2})}^2+\frac{b^2}{a^2}]}dx$$$$=\frac{1}{2}Re\left\{e^{-\frac{b^2}{a^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{(ax+\frac{ib}{a})}^2}dx\right\}$$$$=\frac{1}{2}Re\left\{e^{-\frac{b^2}{a^2}}\frac{1}{a}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-u^2}du\right\}=\frac{\sqrt{\pi }}{2a}{e}^{-\frac{b^2}{a^2}}$$

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