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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Find the extrema: an optimization problem with Lagrange multipliers

Question:
By using Lagrange multipliers, find the extrema of $x^2+2y^2+3z^2$ when $3x+2y+z=-1$.

Answer:
Using a Lagrange multiplier, we define the following function:
\[ F(x,y,z) =  x^2+2y^2+3z^2 - \lambda (3x+2y+z+1) \]
Take each partial derivative.
\[ F_x = 2x-3\lambda = 0  \\
F_y = 4y-2\lambda = 0    \\
F_z = 6z-\lambda = 0   \\
F_{\lambda} = 3x+2y+z+1 = 0   \]
They have four unknowns and four equations. Solve for each variable.
\[ x=-9/34  \\
  y=-3/34  \\
  z=1/34  \\
  \lambda = -6/34 \]
If you plug $x$, $y$, and $z$, we obtain $F = 3/34$ as the extremum.

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Finding the extrema; an optimization problem

Question:
Find the extrema of $x^2+2y^2+3z^2$ when $3x+2y+z=-1$.

Answer:
From the condition, we can solve for $z$.
\[ z = -(3x+2y+1)\]
Thus, we can define a function.
\[ F(x,y) = x^2+2y^2+3(-(3x+2y+1)) \]
It is supposed to find the extrema of this function. Take the partial derivatives.
\[ \frac{\partial F(x,y)}{\partial x} = 56x+36y+18,  \\
  \frac{\partial F(x,y)}{\partial y} = 36x+28y+12,   \\
   \frac{\partial^2 F(x,y)}{\partial x^2} = 56,  \\
    \frac{\partial^2 F(x,y)}{\partial y^2} = 28,  \\
      \frac{\partial^2 F(x,y)}{\partial x \partial y} = 36  \]
The value of the extremum is found when $\frac{\partial F(x,y)}{\partial x} = 0$ and $\frac{\partial F(x,y)}{\partial y} = 0$. Namely,
\[ x = -\frac{9}{34}  \\
   y = -\frac{3}{34}  \]
In order to find if this is maximum or minimum, calculate $D = F_{xx}F_{yy}-F^2_{xy}$. Then, use the following condition:
If $D>0$ and $F_{xx}>0$, then it is a minimum.
If $D>0$ and $F_{xx}<0$, then it is a maximum.
If $D<0$, then it is not a extremum.
If $D=0$, then it cannot be determined in this method.
For this problem, $D = 272>0$ and $F_{xx}=56>0$, so $F(-9/34,-3/34) = 3/34$ is the minimum.

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