# Homework Help: Impossible Integral

1. Feb 26, 2010

Is it possible to integrate $$\int \frac{1}{1+e^{-x^2}} dx$$

2. Feb 26, 2010

### tiny-tim

Well, it's not "possible" to integrate the easier indefinite integral ∫ e-x2 dx , so I shouldn't think so.

3. Feb 27, 2010

### g_edgar

Any continuous function is integrable on any bounded interval. So this function is. Its indefinite integral exists. But (as noted) the answer is not an elementary function.

4. Mar 4, 2010

### misnomered

Note: This is my first post, so if there is some sort of protocol I'm forgetting or some error in my formatting, go easy for me.

My Calculus III professor showed us a method to calculate $$\int_{0}^{\infty} e^{-nx^2}dx$$, using polar coordinates.

Think of $$x$$ and $$y$$ as independent variables and let

$$I = \int_{0}^{\infty} e^{-nx^2}dx = \int_{0}^{\infty} e^{-ny^2}dy.$$

Then

$$I^2 = \left(\int_{0}^{\infty} e^{-nx^2}dx\right)\left(\int_{0}^{\infty} e^{-ny^2}dy\right) = \int_{0}^{\infty}\int_{0}^{\infty} e^{-n(x^2 + y^2)}dxdy$$

Substituting into polar coordinates, we have that $$r^{2} = x^{2} + y^{2}$$ and $$dx dy = r dr d\theta$$ and thus

$$I^2 = \int_{0}^{2\pi}\int_{0}^{\infty} re^{-nr^2}dr d\theta = \left(\int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{\infty} re^{-nr^2}dr\right) = 2\pi\int_{0}^{\infty} re^{-nr^2}dr$$.

We can then take $$u = r^2$$ so that $$\frac{du}{2} = r dr$$ and thus

$$I^{2} = 2\pi\int_{0}^{\infty} re^{-nr^2}dr = \pi\int_{0}^{\infty} e^{-nu}du = \frac{\pi}{n}$$

and thus

$$I = \sqrt{\frac{\pi}{n}}.$$

If you change the bounds of the integral however, calculating the integral becomes very hard.

5. Mar 5, 2010

### sagardip

What i think is that ∫ e-x2 dx is a reduced erf(x) function or error function.
erf(x)=2/(pi^0.5) ∫ e-x2 dx

6. Mar 5, 2010

### Gib Z

You are correct, but the error function is simply what we define to be the integral it is, well, defined by. It's not some sort of theorem or result, the error function is not some different type of function that happens to be equal to this integral.