Question

Question #87a05

Answer 1

`sqrt(pi)/4[e^(1/2)erf(x-i/sqrt(2))+e^(-1/2)erf(x+i/sqrt(2))]+C`

Explanation:

`inte^(-x^2)cos(sqrt(2)x)dx`

This function cannot be written in terms of typical elementary functions, in that regard we have to use a special function. More specifically we need to use the Gaussian Error function which is defined as follows:

`erf(x) = 2/sqrt(pi)inte^(-x^2)dx`

The definition is generalized for coefficients in front of the exponent like so:

`(erf(sqrt(a)x))/sqrt(a) = 2/sqrt(pi)inte^(-ax^2)dx`

The graph of the Gaussian error function looks like this:

Several techniques can be used to deduce the behavior of the `erf(x)` function from the Guassian integral. One such method is to find the Taylor series of `e^(-x^2)` and integrate that term by term.

This is now diverting from the main question, if you wish to find out more about the Gaussian Error function then follow the link:
http://mathworld.wolfram.com/Erf.html

So now that we have been introduced to the Gaussian Error function we can answer the question, start by using:

`cos(ax) = (e^(iax)+e^(-iax))/2` so we may re-write the integral as:

`inte^(-x^2)cos(sqrt(2)x)dx =inte^(-x^2)(e^(isqrt(2)x)+e^(-isqrt(2)x))/2dx`

Now combine the exponents:

`=1/2inte^(-x^2+isqrt(2)x)+e^(-x^2-isqrt(2)x)dx`

To continue we need to complete the square of the exponents:

`-x^2+isqrt(2)x=-(x-isqrt(2)/2)^2-(-isqrt(2)/2)^2`

`=-(x-i/sqrt(2))^2+1/2`

And

`-x^2-isqrt(2)x=-(x+isqrt(2)/2)^2-(+isqrt(2)/2)^2`

`=-(x+i/sqrt(2))^2-1/2`

So; our integral now becomes:

`1/2inte^(-(x-i/sqrt(2))^2+1/2)+e^(-(x+i/sqrt(2))^2-1/2)dx`

`=e^(1/2)/2inte^(-(x-i/sqrt(2))^2)dx+e^(-1/2)/2inte^(-(x+i/sqrt(2))^2)dx`

Now substitute `u=x-i/sqrt2-> du=dx` for the first integral.

Substitute: `v = x+i/sqrt(2)-> dv=dx` for the second integral.

So we now have:

`=e^(1/2)/2inte^(-u^2)du+e^(-1/2)/2inte^(-v^2)dv`

Now, using our definition for the `erf(x)` function we can integrate this to get:

`(e^(1/2)sqrt(pi))/4erf(u)+(e^(-1/2)sqrt(pi))/4erf(v)+C`

Now we can reverse the substitution:

`sqrt(pi)/4[e^(1/2)erf(x-i/sqrt(2))+e^(-1/2)erf(x+i/sqrt(2))]+C`