# Converging integrals

• May 4th 2008, 09:56 AM
matty888
Converging integrals
Show whether or not each of the following integrals converge:

• May 4th 2008, 10:23 AM
Peritus
$\displaystyle \int\limits_1^\infty {e^{ - \left( {3x^2 + 2} \right)} dx = e^{ - 2} } \int\limits_1^\infty {e^{ - 3x^2 } dx}$

now it's easy to verify that:

$\displaystyle e^{ - 3x^2 } \leqslant e^{ - 3x} \quad \forall 1 \leqslant x$

thus:

$\displaystyle \int\limits_1^\infty {e^{ - 3x^2 } } dx < \int\limits_1^\infty {e^{ - 3x} } dx\quad \forall 1 \leqslant x$

now:

$\displaystyle \int\limits_1^\infty {e^{ - 3x} } dx = \left. { - \frac{1} {3}e^{ - 3x} } \right|_1^\infty = \frac{1} {3}e^{ - 3} < \infty$

thus according to the comparison test we conclude that the integral converges.

2.

note that the integrand is an even function, thus:

$\displaystyle \int\limits_{ - \infty }^\infty {e^{ - x^2 } dx} = 2\int\limits_0^\infty {e^{ - x^2 } dx}$

now following the lines of the previous exercise we note that:

$\displaystyle e^{ - x^2 } \leqslant e^{ - x} \quad \forall x \geqslant 0$

....
• May 4th 2008, 10:40 AM
Krizalid
Quote:

Originally Posted by matty888

Since $\displaystyle e^x\ge x+1,\,\forall\,x\in\mathbb R$ we have $\displaystyle e^{x^{2}}\ge x^{2}+1\implies e^{-x^{2}}\le \frac{1}{x^{2}+1}.$

Hence $\displaystyle \int_{-\infty }^{+\infty }{e^{-x^{2}}\,dx}\le \int_{-\infty }^{+\infty }{\frac{1}{1+x^{2}}\,dx}=\pi .$ The given integral converges.
• May 4th 2008, 10:48 AM
Peritus
Quote:

Originally Posted by Krizalid
Since $\displaystyle e^x\ge x+1,\,\forall\,x\in\mathbb R$ we have $\displaystyle e^{x^{2}}\ge x^{2}+1\implies e^{-x^{2}}\le \frac{1}{x^{2}+1}.$

Hence $\displaystyle \int_{-\infty }^{+\infty }{e^{-x^{2}}\,dx}\le \int_{-\infty }^{+\infty }{\frac{1}{1+x^{2}}\,dx}=\pi .$ The given integral converges.

nice (Yes)