Help with a definite integral property

[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

 

[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

 

[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

 

[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

 

[pamparana]

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

Cheers,
Luca

 

[LCKurtz]

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

e^{-x^2}

 

[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

 

[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

 

[fourier jr]

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