(응용)수학 [(응용)수학] Why does ∫∞0e−t2cos(at)dt=12π−−√e−a2/4∫0∞e−t2cos⁡(at)dt=12πe−a2/4 hold?

Let I(a)$I(a)$ denote the parameterized improper integral:

I(a)=∫∞0e−t2cos(at)dt$\begin{array}{}& I(a)={\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-t^2}\cos (at)dt}\end{array}$

Differentiating I(a)$I(a)$ yields:

I′(a)=∫∞0∂∂a(e−t2cos(at))dt$\begin{array}{}& I^{\prime }(a)={\displaystyle {\int }_0^{\mathrm{\infty }}{\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }a}}(e^{-t^2}\cos (at))dt}\end{array}$

=−∫∞0te−t2sin(at)dt$\begin{array}{}& =-{\displaystyle {\int }_0^{\mathrm{\infty }}t{e}^{-t^2}\sin (at)dt}\end{array}$

Now for the remaining integral, applying integration by parts with u=sin(at)$u=\sin (at)$ and dv=te−t2dt$dv=t{e}^{-t^2}dt$ yields:

I′(a)=IBP12e−t2sin(at)∣∣∣∞0−a2∫∞0e−t2cos(at)dt$\begin{array}{}& I^{\prime }(a)\stackrel{\text{IBP}}{=}{{\displaystyle \frac{1}{2}}{e}^{-t^2}\sin (at)|}_0^{\mathrm{\infty }}-{\displaystyle \frac{a}{2}}{\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-t^2}\cos (at)dt}\end{array}$

=−a2∫∞0e−t2cos(at)dt=−a2I(a)$\begin{array}{}& =-{\displaystyle \frac{a}{2}}{\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-t^2}\cos (at)dt=-{\displaystyle \frac{a}{2}}I(a)}\end{array}$

Hence we obtain the first order linear differential equation:

I′(a)=−a2I(a)$\begin{array}{}& I^{\prime }(a)=-{\displaystyle \frac{a}{2}}I(a)\end{array}$

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whose solution is:

I(a)=ce−a2/4$\begin{array}{}& I(a)=c{e}^{-a^2/4}\end{array}$

For some constant c∈R$c\in \mathbb{R}$.

As I(0)=∫∞0e−t2dt=π−−√/2$I(0)={\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt=\sqrt{\pi }/2$ (known as Guassian integral) we get:

I(a)=π−−√2e−a2/4$\begin{array}{}& \boxed{{\displaystyle I(a)={\displaystyle \frac{\sqrt{\pi }}{2}}{e}^{-a^2/4}}}\end{array}$

as required.