# Help with a definite integral property

1. Jan 7, 2010

### pamparana

Hello,

Saw this expression:

$$^{inf}_{-inf} \int e^{x^{2}} = 2 \times ^{inf}_{+0} \int e^{x^{2}}$$

I am unable to understand this change of base. You can change from -inf to 0 and 0 to +inf but do not see how that equals to 2 * the integral from 0 to infinity.

I hope someone will help me see the light :)

Thanks,

Luc

2. Jan 7, 2010

### fourier jr

it's because $$e^{x^2}$$ is an even function, that is, it satisfies the equation f(-x) = f(x). if you look at the graph of the function it's symmetric about the y-axis

3. Jan 7, 2010

### pamparana

That makes sense. Thanks for that. I still have one confusion though:

$$\int^{\infty}_{-\infty} e^{x^{2}} = \int^{0}_{-\infty} e^{x^{2}} + \int^{\infty}_{0} e^{x^{2}}$$

Now using the function symmetry we have:

$$\int^{\infty}_{-\infty} e^{x^{2}} = \int^{-\infty}_{0} e^{x^{2}} + \int^{\infty}_{0} e^{x^{2}}$$

However, the limits are still messed up in the first term, right? The upper limit still has $$-\infty$$

Or have I done this completely wrong?

Thanks,

Luca

4. Jan 7, 2010

### fourier jr

$\int_{-\infty}^{\infty} e^{x^2} dx = \int_{-\infty}^{0} e^{x^2} dx + \int_{0}^{\infty} e^{x^2} dx$ but since $e^{x^2}$ is even, $\int_{-\infty}^{0} e^{x^2} dx = \int_{0}^{\infty} e^{x^2} dx$ so $\int_{-\infty}^{\infty} e^{x^2} dx = 2\int_{0}^{\infty} e^{x^2} dx$

5. Jan 7, 2010

### pamparana

Ahhhh thanks. Did not know you could change the integral limits for an even function like that.

Cheers,
Luca

6. Jan 7, 2010

### LCKurtz

And, of course, all the integrals mentioned above diverge unless the integrand has a negative sign:

$$e^{-x^2}$$

7. Jan 7, 2010

### fourier jr

ARGHHHH i don't know how i missed that! i guess it's too late to delete everything i've added to this thread luckily for me math isn't an exact science

Last edited: Jan 7, 2010
8. Jan 8, 2010

### pamparana

Of course. Checking for convergence or divergence is another business all-together.

However, the change of limits for the integral still holds, right?

Thanks,
Luca

9. Jan 8, 2010

### fourier jr

yes, because $\int_{0}^{\infty} f(x) dx = \lim_{M \rightarrow \infty}\int_{0}^{M} f(x) dx$