Infinite integral with one constant and one variable

Question:
Find the value of
\[
\int^{\infty}_{-\infty}e^{-x^2}\cos 2bxdx
\]

Answer:
Let
\[
\int^{\infty}_{-\infty}f(x,b)dx = 2\int^{\infty}_{0}f(x,b)dx
\]
where $f(x,b)=e^{-x^2}\cos 2bx$. According to the properties of the functions, we can state that $|f(x,b)| \leqq e^{-x^2}$; and $\int^{\infty}_{0}e^{-x^2}$ is finite. Therefore, $\varphi(b)=\int^{\infty}_{0}f(x,b)dx$ is also finite. Now, take the partial derivative of the function $f(x,b)$ with respect to $b$.
\[
f_b(x,b) = -2xe^{-x^2}\sin 2bx
\]
This is continuous for arbitrary $b$ with $x \geqq 0$. Also we know $|f(x,b)| \leqq 2xe^{-x^2}$; and $\int^{\infty}_{0}2xe^{-x^2}$ is finite. Therefore, $\int^{\infty}_{0}f_b(x,b)dx$ is also finite for any $b$. Then,
\begin{eqnarray}
\varphi'(b) &=& \int^{\infty}_{0}f_b(x,b)dx  \\
  &=&  \int^{\infty}_{0}-2xe^{-x^2}\sin 2bxdx  \\
   &=& \left[ e^{-x^2}\sin 2bx \right]^{\infty}_{0} - 2b\int^{\infty}_{0}e^{-x^2}\cos 2bxdx \\
   &=& 0-2b\varphi(b)
\end{eqnarray}
We have the following:
\begin{equation}
\frac{\varphi'(b)}{\varphi(b)}=-2b
\end{equation}
Then, integrate both sides.
\begin{eqnarray}
\ln|\varphi(b)| &=& -b^2 + C  \\
\varphi(b) &=& Ce^{-b^2}
\end{eqnarray}
From the initial condition,
\begin{equation}
\varphi(0) = \int^{\infty}_{0}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}=C
\end{equation}
Thus,
\begin{equation}
\varphi(b)=\frac{\sqrt{\pi}}{2}e^{-b^2}
\end{equation}
Therefore,
\begin{equation}
\int^{\infty}_{-\infty}e^{-x^2}\cos 2bxdx = \sqrt{\pi}e^{-b^2}
\end{equation}

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