Question

What is `int_1^oo e^(-x^2) dx+int_-oo^0 e^(-x^2)dx`?

Answer 1

`int_(1)^(oo) \ e^(-x^2) \ dx + int_(-oo)^(0) \ e^(-x^2) \ dx = 1.025629718 ...`

Explanation:

Let:

`I = int_(1)^(oo) \ e^(-x^2) \ dx + int_(-oo)^(0) \ e^(-x^2) \ dx`

Then;

`I + int_(0)^(1) \ e^(-x^2) \ dx= int_(-oo)^(oo) \ e^(-x^2) \ dx`

Now the integral of `e^(-x^2)` does not have an elementary solution, but the RHS definte integral has a surprising result (discovered by Carl Guass) of `sqrt(pi)`, and for the LHS we use what is known as the error function `erf(x)`.

`erf(x) = 2/sqrt(pi) \ int_0^x e^(-t^2) \ dt`

So we can write:

`I + sqrt(pi)/2erf(1) = sqrt(pi)`
`:. I = sqrt(pi) - sqrt(pi)/2 \ erf(1)`

And the values of the error function can be looked up in tables:

`erf(1) = 0.842700792 ...`

So we have:

`:. I = 1.025629718 ...`