A convergence and divergence test and a couple integrals

[Bob Busby]

Here are five separate problems.

Show that the series 1/3^(ln(n)) converges and that the series 1/2^(ln(n)) diverges.

integral (sqrt(x)*e^-sqrt(x)) dx

integral (x/(sqrt(x-1)+2)) dx

integral (1/(2+sin(x)+cos(x)

integral (1-cos(x))^(5/2)) dx

There are no other relevant equations.

My attempts at solution lead nowhere and are too long-winded to post. I believe I am just missing one important premise I need to solve these. Any help would be appreciated.

 

[hgfalling]

Here are some top-level hints:

OK, for the series, try using the integral test.

For integral #1, make the obvious u substitution, then integrate by parts twice.
For integral #2, make the obvious u substitution, then grind through the rational polynomials integrations.
For integral #3, use the tangent half-angle substitution.
For integral #4, do two u substitutions.

 

[Bob Busby]

Thanks for the help.

I made a slight parentheses mistake with the last two integrals. They should be:

integral (1/(2+sin(x)+cos(x)) dx

integral ((1-cos(x))^(5/2)) dx

Does this change your hints at all? I tried manipulating the tangent half-angle formula, but I just turns into mess.

With the 4th integral it seems that no matter what I substitute I end up with nasty, unmanageable square roots when I try to get rid of the sin(x) that comes with the dx.

I'd be grateful for any more help.

 

[hgfalling]

For the third integral, we make the substitution t = \tan \frac{x}{2}. Then:

\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1 - t^2}{1 + t^2}, dx = \frac{2}{1 + t^2} dt.

OK, so plug all that in and simplify; you wind up with this:

\int \frac{2 dt}{3 + t^2 + 2t}

Now just do a little rearranging/substitution to get it into a familiar inverse trig form, and everything is spiffy.

For the fourth integral, you have:

\int (1 - \cos x)^{\frac{5}{2}} dx

Make the substitution u = 1 - \cos x. So du = \sin x dx.

Now there isn't a \sin x handy in your original integrand, but that's ok.

Suppose we took 2 - u, which is 1 + \cos x , and we multiplied it by u, which is 1 - \cos x . That gives us 1 - \cos^2 x, which is conveniently equal to \sin^2 x .

So \sin x = \sqrt{u(2-u)}. Since this is part of du, we need to divide the integrand by it. We conveniently have a factor of \sqrt{u} already in the integrand, so we just divide out the \sqrt{2-u} term, which gives us:

\int \frac{u^2}{\sqrt{2-u}} du

I'm sure you can take it from here.

 

[Bob Busby]

Awesome. Thanks for the help!

I have a question about the series, though. I can show that the series is always decreasing but I don't know how to integrate 1/3^(ln(n)). I tried using a u-substitution and I ended up with the integral of (e^u/3^u) du which I don't know how to deal with either.

Somebody told me I should use the ratio test, but it ended up being the limit as n goes to infinity of 3^((n)-(n+1)) which is 1 so the result told me nothing. She also said to compare the values of ln(2) and ln(3), but those aren't even in the series. I'm still lost with this problem.

 

[hgfalling]

I don't see how you can use the ratio test for this one, although I may be missing something.

To integrate \int a^{- \ln x} dx, make the substitution u = a^{- \ln x}. Now du will have an x in the denominator by the chain rule, so you'll need to find a way to express x in terms of u (hint: a^{\log_a x} = x). Once you do one of the integrals, the other one is just the same. Then simply show that the a = 3 integral converges, while the a=2 integral diverges.

Then, for fun, make a general statement about the series convergence for any value of a.

 

[Bob Busby]

I ended up with the integral of -(u*x)/(ln(3)*3^-ln(x)) du

I solved for x and found x = e^log_3(u)

After substituting I got -(e^log_3(u)*u)/(ln(3)*3^-log_3(u)) du which baffles me.
If I wanted to I could change 3^-log_3(u) into 1/u but I still don't see where to go from there.

 

[hgfalling]

\int a^{- \ln x} dx
Make the substitution:
u = a^{- \ln x}
du = \frac{(\ln a) (a^{- \ln x})}{x} dx = \frac{(\ln a) u}{x} dx

Find x in terms of u:
\log_a u = - \ln x

\frac{\ln u}{\ln a} = -\ln x

x = -u^{\frac{1}{\ln a}}

So now:
du = \frac{(\ln a) (a^{- \ln x})}{x} dx = \frac{(\ln a) u}{-u^{\frac{1}{\ln a}}} dx

du = -(\ln a) u^{1 - \frac{1}{\ln a}} dx

OK, now back into the original integral. To make the du work, we need to divide the original integral by -(\ln a) u^{1 - \frac{1}{\ln a}}. So:

\int a^{- \ln x} dx = - \int \frac{u}{(\ln a)(u^{1 - \frac{1}{\ln a}})} du = - \int \frac{u^{\frac{1}{\ln a}}}{\ln a} du

OK, that's the tricky part. The rest is pretty easy.

 

[Bob Busby]

Sweet. Then the indefinite integral should be:

(a^(-ln(x)*1/(ln(3))+1))/(ln(a) +1)

The bad news is that evaluating this at infinity and 1 (for the integral test) gives -.59062 for a = 2 and -.47651 for a =3 which means that both of the series converge, which can not be true as the problem tells me that one converges and one diverges.

What's going on?

 

[Dick]

(1/2)^(log(x)) is actually x^p for some constant p. Can you find that p? (1/2)^(log(x))=e^(log(1/2)*log(x))=(e^(log(x)))^log(1/2). What do you know about the convergence of p-series?

 

[hgfalling]

From here it goes:
- \int \frac{u^{\frac{1}{\ln a}}}{\ln a} du = - \frac{u^{(1 -\frac{1}{\ln a})}}{(1 - \frac{1}{\ln a}) \ln a}

Substituting back for u:

-\frac{{(a^{-\ln x})}^{(1 -\frac{1}{\ln a})}}{\ln a - 1} = \frac{x a^{- \ln x}}{1 - \ln a} (+C)

 

[Dick]

From here it goes:
- \int \frac{u^{\frac{1}{\ln a}}}{\ln a} du = - \frac{u^{(1 -\frac{1}{\ln a})}}{(1 - \frac{1}{\ln a}) \ln a}

Substituting back for u:

-\frac{{(a^{-\ln x})}^{(1 -\frac{1}{\ln a})}}{\ln a - 1} = \frac{x a^{- \ln x}}{1 - \ln a} (+C)

Aren't you making this a bit harder than it needs to be? a^(ln(x))=x^(ln(a)), isn't it?

 

[hgfalling]

Yes, much harder. :(

OP, pay attention to what Dick says.

 

[Bob Busby]

(1/2)^(log(x)) is actually x^p for some constant p. Can you find that p? (1/2)^(log(x))=e^(log(1/2)*log(x))=(e^(log(x)))^log(1/2). What do you know about the convergence of p-series?

That's so beautiful yet simple. That was the clue I needed.

Thanks to both of you for your help.

 

[gomunkul51]

Bob Busby:

you can use Cauchy's Condensation Test for the first part of question 1:
f(n) =< (2^n)*f(2^n)

Also can be done by the Ye'Olde p Test :)
~ 1/(n^1.1)