# diverging or converging?

• Jan 28th 2009, 05:24 PM
razorfever
diverging or converging?
determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 0 to infinity: xe^-(x^2) dx

integral from 0 to 1: xln(x) dx

integral from 1 to infinity: (ln t) / (t^2) dt

integral from 0 to 2: 1 / (t-1) dt

i have solved the first integral and i got the answer as 1/2 but how do i show if it is diverging or converging?
• Jan 28th 2009, 05:30 PM
Mush
Quote:

Originally Posted by razorfever
determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 0 to infinity: xe^-(x^2) dx

integral from 0 to 1: xln(x) dx

integral from 1 to infinity: (ln t) / (t^2) dt

integral from 0 to 2: 1 / (t-1) dt

i have solved the first integral and i got the answer as 1/2 but how do i show if it is diverging or converging?

The integral converges if it's value is finite. If it's value is infinite, it is divergent.

$\displaystyle \int_0^1 {x\ln(x) dx} = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x^2}{2} \times \frac{1}{x} dx}$

$\displaystyle = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x}{2} dx}$

$\displaystyle = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \bigg[{\frac{x^2}{4}}\bigg]^1_0$

This integral is divergent because when you plug in the limits in the first expression, you end up with the term $\displaystyle \ln(0)$. This is not defined, and hence the integral has no finite value, it does not converge.

Try the other 2 and let us know what results you obtain.
• Jan 28th 2009, 07:17 PM
Krizalid
Quote:

Originally Posted by Mush

The integral converges if it's value is finite. If it's value is infinite, it is divergent.

$\displaystyle \int_0^1 {x\ln(x) dx} = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x^2}{2} \times \frac{1}{x} dx}$

$\displaystyle = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x}{2} dx}$

$\displaystyle = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \bigg[{\frac{x^2}{4}}\bigg]^1_0$

This integral is divergent

Since $\displaystyle \underset{x\to 0}{\mathop{\lim }}\,x\ln x=0,$ then $\displaystyle x\ln x$ is continuous on $\displaystyle (0,1],$ hence integrable on $\displaystyle [0,1].$
• Jan 28th 2009, 08:36 PM
razorfever
so does that mean its diverging or converging, now im totally confused??
• Jan 28th 2009, 08:53 PM
Krizalid
Convergent.
• Jan 28th 2009, 09:23 PM
razorfever
hah so i was right ... thanks for clearing that up
• Jan 29th 2009, 06:26 AM
Krizalid
Quote:

Originally Posted by razorfever

determine whether the integral diverges or converges
if it converges then evaluate the integral

integral from 0 to infinity: xe^-(x^2) dx

i have solved the first integral and i got the answer as 1/2 but how do i show if it is diverging or converging?

$\displaystyle \int_{0}^{\infty }{x{{e}^{-{{x}^{2}}}}\,dx}=\int_{0}^{1}{x{{e}^{-{{x}^{2}}}}\,dx}+\int_{1}^{\infty }{x{{e}^{-{{x}^{2}}}}\,dx}.$ Let's just worry about the second piece since the first one is integrable on $\displaystyle [0,1]$ for being a continuous function on that interval. Now, for $\displaystyle x\ge1$ we have $\displaystyle \frac{x}{{{e}^{{{x}^{2}}}}}\le \frac{x}{{{e}^{x}}}=x{{e}^{-x}},$ hence by direct comparison test your integral converges because $\displaystyle \int_{1}^{\infty }{x{{e}^{-x}}\,dx},$ does.

Second and third question were already answered, and for your last question, you just need to split that integral into two ones. I mean, $\displaystyle \int_{0}^{2}{\frac{dt}{t-1}}=\int_{0}^{1}{\frac{dt}{t-1}}+\int_{1}^{2}{\frac{dt}{t-1}}.$

And there's no much here, these are clearly divergent integrals, so your integral diverges.