# Thread: Improper Integrals, Convergence, Sequence Convergence? Ahh!

1. ## Improper Integrals, Convergence, Sequence Convergence? Ahh!

Hey there! Having problems with a few of the exercises on some homework!

Do the integrals converge or diverge? If they converge, what is their value?
a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
b) integral from 0 to 2 of ln(x)dx
c)integral from zero to infinity of dx/(2x-1)

Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

Thank you!
My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??

2. Originally Posted by matt.qmar
Hey there! Having problems with a few of the exercises on some homework!

Do the integrals converge or diverge? If they converge, what is their value?
a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
b) integral from 0 to 2 of ln(x)dx
c)integral from zero to infinity of dx/(2x-1)

Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

Thank you!
My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx

$\int\limits_1^\infty {\frac{{{x^3} + x}}
{{{x^4} + 2{x^2} + 2}}dx} = \frac{1}
{4}\int\limits_1^\infty {\frac{{4{x^3} + 4x}}
{{{x^4} + 2{x^2} + 2}}dx} = \frac{1}
{4}\int\limits_1^\infty {\frac{{d\left( {{x^4} + 2{x^2} + 2} \right)}}
{{{x^4} + 2{x^2} + 2}}} =$

$= \left. {\frac{1}{4}\ln \left( {{x^4} + 2{x^2} + 2} \right)} \right|_1^\infty = \frac{1}{4}\mathop {\lim }\limits_{x \to \infty } \ln \left( {{x^4} + 2{x^2} + 2} \right) - \frac{{\ln 5}}{4} = \infty .$

So, this integral is divergent.

3. ## One additional Question

Also, in part a)
Does the integral still diverge if the denominator is (x^4 + 2*x^2 + 2)^1/2?

Thank you so much!

4. Originally Posted by matt.qmar
Hey there! Having problems with a few of the exercises on some homework!

Do the integrals converge or diverge? If they converge, what is their value?
a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
b) integral from 0 to 2 of ln(x)dx
c)integral from zero to infinity of dx/(2x-1)

Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

Thank you!
My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
b) integral from 0 to 2 of ln(x)dx

$\int\limits_0^2 {\ln xdx} = \left. {x\ln x} \right|_0^2 - \int\limits_0^2 {dx} = \left. {\left( {x\ln x - x} \right)} \right|_0^2 = 2\left( {\ln 2 - 1} \right) - \mathop {\lim }\limits_{x \to 0} \left( {x\ln x - x} \right).$

$\mathop {\lim }\limits_{x \to 0} \left( {x\ln x - x} \right) = \mathop {\lim }\limits_{x \to 0} x\ln x - \mathop {\lim }\limits_{x \to 0} x = \mathop {\lim }\limits_{x \to 0} x\ln x = \left\{ \begin{gathered}
x = \frac{1}
{t}, \hfill \\
x \to 0, \hfill \\
t \to \infty \hfill \\
\end{gathered} \right\} =$

$= \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \frac{1}
{t}}}
{t} = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{d}
{{dt}}\ln \frac{1}
{t}}}
{{\frac{{dt}}
{{dt}}}} = - \mathop {\lim }\limits_{t \to \infty } \frac{1}
{t} = 0.$

Thus, the integral $\int\limits_0^2 {\ln xdx}$ converges, because it has a finite value and is equal to $2\left( {\ln 2 - 1} \right).$

5. ## Progress!

I think I've solved Question (2)!
let N=3/(2*epsilon)
We can show abs(a sub n - (5/2)) < epsilon?
(this is just the beginning and end of the proof of course!)

6. Originally Posted by matt.qmar
Also, in part a)
Does the integral still diverge if the denominator is (x^4 + 2*x^2 + 2)^1/2?

Thank you so much!
Similarly

$\int\limits_1^\infty {\frac{{{x^3} + x}}
{{\sqrt {{x^4} + 2{x^2} + 2} }}\,dx} = \frac{1}
{4}\int\limits_1^\infty {\frac{{4{x^3} + 4x}}
{{\sqrt {{x^4} + 2{x^2} + 2} }}\,dx} = \frac{1}
{4}\int\limits_1^\infty {\frac{{d\left( {{x^4} + 2{x^2} + 2} \right)}}
{{\sqrt {{x^4} + 2{x^2} + 2} }}} =$

$= \left. {\frac{1}{2}\sqrt {{x^4} + 2{x^2} + 2} } \right|_1^\infty = \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \sqrt {{x^4} + 2{x^2} + 2} - \frac{{\sqrt 5 }}{2} = \infty .$

7. Originally Posted by matt.qmar
Hey there! Having problems with a few of the exercises on some homework!

Do the integrals converge or diverge? If they converge, what is their value?
a) integral from 1 to infinity of ((x^3 + x)/((x^4 + 2x^2 + 2)))dx
b) integral from 0 to 2 of ln(x)dx
c)integral from zero to infinity of dx/(2x-1)

Also, prove, using the formal definition, that the sequence (5n+3)/(2n) from n=1 to infinity, converges.

Thank you!
My leads are, I am pretty sure b) and c) diverge, but I am not sure exactly! Also, I know the E-D def'n for the sequence but not how to apply it to convergence??
c) integral from zero to infinity of dx/(2x-1)

$\int\limits_0^\infty {\frac{{dx}}
{{2x - 1}}} = \frac{1}
{2}\int\limits_0^\infty {\frac{{d\left( {2x - 1} \right)}}
{{2x - 1}}} = \left. {\frac{1}
{2}\ln \left| {2x - 1} \right|} \right|_0^\infty = \frac{1}
{2}\mathop {\lim }\limits_{x \to \infty } \ln \left| {2x - 1} \right| = \infty .$

So, this integral is divergent.