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Consider the ODE $y'+xy=1$, the solution is:

$$y=Ce^{-x^2/2}+e^{-x^2/2}\int_0^xe^{t^2/2}dt$$ With Laplace trasform, another solution is found as:

$$y=C\int_0^{\infty}e^{-t^2/2}\cos(xt)dt+\int_0^{\infty}e^{-t^2/2}\sin(xt)dt$$ Since the ODE has a unique solution, these two solutions must be identical. How to prove these two integral equalities?

\begin{align} &\int_0^{\infty}e^{-t^2/2}\cos(xt)dt=\sqrt\frac{\pi}2e^{-x^2/2} \\ &\int_0^{\infty}e^{-t^2/2}\sin(xt)dt=e^{-x^2/2}\int_0^xe^{t^2/2}dt \end{align} I got a proof with the Taylor series of $\cos(xt)$ and $\sin(xt)$. Is there any other proof without using the Taylor series?

Nanhui Lee
  • 856

1 Answers1

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Thanks to the guys who comment for the solution hints. I have derived a proof with Gaussian integral erf(x) and erfi(x) functions.

\begin{align} &\int_0^{\infty}e^{-t^2/2}\cos(xt)dt+i\int_0^{\infty}e^{-t^2/2}\sin(xt)dt\\ &=\int_0^{\infty}e^{-t^2/2}e^{ixt}dt=\int_0^{\infty}e^{-t^2/2+ixt}dt\\ &=\int_0^{\infty}e^{-x^2/2}e^{-(t^2-2ixt-x^2)/2}dt=e^{-x^2/2}\int_0^{\infty}e^{-(t-ix)^2/2}dt\\ &=e^{-x^2/2}\left [\sqrt\frac{\pi}2\operatorname{erf}\left ( \frac{t-ix}{\sqrt2} \right ) \right ]_{t=0}^{t=\infty}\\ &=\sqrt\frac{\pi}2e^{-x^2/2}\left [\operatorname{erf}\left ( \frac{\infty-ix}{\sqrt2}\right )-\operatorname{erf}\left ( \frac{-ix}{\sqrt2}\right )\right]\\ &=\sqrt\frac{\pi}2e^{-x^2/2}\left [1+i\operatorname{erfi}\left ( \frac x{\sqrt2}\right )\right]\\ &=\sqrt\frac{\pi}2e^{-x^2/2}+ie^{-x^2/2}\int_0^xe^{t^2/2}dt\\ \\ &Note: \operatorname{erf}(\infty)=1,\quad\operatorname{erf}(-x)=-\operatorname{erf}(x),\quad\operatorname{erf}(ix)=i\operatorname{erfi}(x) \end{align} So these two equalities are proved. \begin{align} &\int_0^{\infty}e^{-t^2/2}\cos(xt)dt=\sqrt\frac{\pi}2e^{-x^2/2} \\ &\int_0^{\infty}e^{-t^2/2}\sin(xt)dt=e^{-x^2/2}\int_0^xe^{t^2/2}dt \end{align}

Nanhui Lee
  • 856