calculuscompleting-the-squareintegration
How can I simplify
$ 1 – \frac{1}{\sqrt{2\pi}\sigma}\int^{\infty}_{-\infty}e^{-kw}e^{\frac{-(w-\mu)^{2}}{2\sigma^{2}}}dw$
using the method of completing the square? I know that the answer must be
$ 1 – e^{- (k\mu – \frac{k^2\sigma^2}{2})}$
and that it must therefore probably involve the Gaussian integral at some point, but I just can't see what the intermediate steps would be.
"I just don't understand why we can complete the square."
The best way to see why it works, is to see a pictorial representation of what is going on, see:
The $x^2$ term is a square with both sides of length $x$. The $bx$ term is a rectangle with one side of length $x$ and one side of length $b$. When you divide by two you are splitting the $bx$ rectangle in two parts with still a side of length $x$ and the other side is now $\frac b2$.
When you place both rectangles as shown on the picture there is only a small square with sides of length $\frac b2$ missing. This small square has area $\frac {b^2}{4}$ and it is this part that you need to "complete the square".
If you have a term $a\ne 1$ in $ax^2+bx+c$ then you have to first "get it off" to get a square with both sides of length $x$. In that case the rectangle you get still has a side of length $x$ but its other side is now $\frac ba$ and this makes the area of the tiny missing square equal to $\frac {b^2}{4a}$.
$$\int { \frac { dx }{ 4x^{ 2 }-4x-3 } } =\int { \frac { dx }{ \left( 2x+1 \right) \left( 2x-3 \right) } } =-\frac { 1 }{ 4 } \int { \left[ \frac { 1 }{ 2x+1 } -\frac { 1 }{ 2x-3 } \right] dx } \\ =-\frac { 1 }{ 4 } \left[ \int { \frac { dx }{ 2x+1 } } -\int { \frac { dx }{ 2x-1 } } \right] =-\frac { 1 }{ 8 } \left[ \int { \frac { d\left( 2x+1 \right) }{ 2x+1 } } -\int { \frac { d\left( 2x-1 \right) }{ 2x-1 } } \right] =\\ =-\frac { 1 }{ 8 } \left[ \ln { \left| 2x+1 \right| -\ln { \left| 2x-1 \right| } } \right] =-\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 } \right| } +C=\\ \\ or\\ -\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 } \right| } +C=-\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 2x-1 } \right) } =\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 1-2x } \right) } \\ \\ $$ Obviously,here should be $\tanh ^{ -1 }{ x } $ not $\tan ^{ -1 }{ x } $
Best Answer
Let us foccus on the exponents. Multiplying out the squared brackets and adding the exponent $-kw$.
$$\frac{-(w^2-2w\mu+\mu^2)}{2\sigma^2}-kw\frac{2\sigma^2}{2\sigma^2}$$
$$\frac{-w^2+2w\mu-\mu^2}{2\sigma^2}-kw\frac{2\sigma^2}{2\sigma^2}$$
Collecting terms with the variable $w$
$$\frac{-w^2+(2\mu-2\sigma^2k)w}{2\sigma^2}-\frac{\mu^2}{2\sigma^2}$$
Completing the square
$$\frac{-w^2+(2\mu-2\sigma^2k)w-(\mu-\sigma^2k)^2}{2\sigma^2}-\frac{\mu^2-(\mu-\sigma^2k)^2}{2\sigma^2}$$
$$\frac{-(w-(\mu-\sigma^2k))^2}{2\sigma^2}-\frac{\mu^2-(\mu-\sigma^2k)^2}{2\sigma^2}$$
$$\frac{-(w-(\mu-\sigma^2k))^2}{2\sigma^2}-\frac{\mu^2}{2\sigma^2}+\frac{\mu^2-2\mu\sigma^2k+\sigma^4k^2}{2\sigma^2}$$
$$\frac{-(w-(\mu-\sigma^2k))^2}{2\sigma^2}\underbrace{-\frac{\mu^2}{2\sigma^2}+\frac{\mu^2}{2\sigma^2}}_{=0}+\frac{-2\mu\sigma^2k+\sigma^4k^2}{2\sigma^2}$$
Cancelling out $\sigma^2$ at the last fraction
$$\frac{-(w-(\mu-\sigma^2k))^2}{2\sigma^2}+\frac{-2\mu k+\sigma^2k^2}{2}$$
$$\frac{-(w-(\mu-\sigma^2k))^2}{2\sigma^2}-\mu k+\frac{\sigma^2k^2}{2}$$
Now we integrate
$$e^{-\mu k+\frac{\sigma^2k^2}{2}}\cdot \underbrace{\frac{1}{\sqrt{2\pi}\sigma}\cdot \int_{-\infty}^{\infty} e^{\frac{-(w-(\mu-\sigma^2k))^2}{2\sigma^2}} \, dw}_{=1}$$