Question

Use the Comparison Theorem (CT) to determine whether this integral is convergent or divergent
[latex]
\[\int\limits_{1}^{\infty}\frac{33(2+e^{-x})}{x}dx\]

Answer 1

Ahh!Messed it up. One sec.

Answer 2

@FoolForMath

Answer 3

\[\int\limits_{1}^{\infty}\frac{33(2+e^{-x})}{x}dx\]

Answer 4

@TuringTest

Answer 5

I said this is divergent, and the comparison I made was with this:
\[\int\limits\limits_{1}^{\infty}\frac{33(2+e^{-x})}{1/x}dx\]

Answer 6

Because:
\[x \ge\frac{1}{x}\]
on [1,infinity]

Answer 7

I want to clarify with you guys to see if it is a right or wrong comparison.

Answer 8

So basically, what I did was I solved for the one with "1/x"

Answer 9

I don't think that's a good comparison, but I'm struggling to see why we even need the comparison test...

Answer 10

Because it asks to use the CT.

Answer 11

Why is it not a good comparison? Is not the denominator term (1/x) smaller than the term (x)

Answer 12

right, so the integrand is larger for 1/x than with x int the denom
so proving that it diverges with 1/x does not prove that it diverges with x
we need to find a smaller integrand that also diverges

Answer 13

I know:
\[\frac{e^-x}{x}\]

Answer 14

\[\frac{33(2+e^{-x})}{x}\le\frac{33(2+e^{-x})}{\frac1x}\]for \(x\ge1\)

Answer 15

And this has to be smaller.

Answer 16

Would my suggestion work above?

Answer 17

\frac{e^-x}{x}

Answer 18

but\[\frac{e^{-x}}x\le\frac{e^{-x}}{\frac1x}=xe^{-x}\]so by putting 1/x in the denom we are making it bigger

Answer 19

No I meant that see if the following integral is divergent:
\[\int\limits_{1}^{\infty}\frac{e^-x}{x}\]

Answer 20

Because surely, this is smaller than the integral in question.

Answer 21

but that integral is not divergent

Answer 22

Really?

Answer 23

http://www.wolframalpha.com/input/?i=integral+1+to+infty+e%5E%28-x%29%2Fx

Answer 24

not sure if I can prove that... I'd have to think about it....

Answer 25

but your integral is so obviously divergent for other reasons, I just think looking to use the comparison test is overkill

Answer 26

Unfortunately, I need a comparison. I just realized that the one I came up with will be convergent. Ahhhh....

Answer 27

oh now I see how to prove the convergence of\[\int_1^{\infty}\frac{e^{-x}}{x}dx\]not that that helps you...

Answer 28

Let me at least make a quick case for divergence of your integral my way; maybe that will help...

Answer 29

How about:
\[\frac{e-^{x}}{x^2}\]

Answer 30

also convergent

Answer 31

if \[\frac{e^{-x}}x\]converges on [1,infty] then so does \[\frac{e^{-x}}{x^2}\]because the second is less than the first for all x in that interval
you can tell that just by looking

Answer 32

It works!

Answer 33

I think it works...

Answer 34

My view:
in your integral\[\int\limits_{1}^{\infty}\frac{33(2+e^{-x})}{x}dx=33[2\int\limits_{1}^{\infty}\frac1xdx+\int\limits_{1}^{\infty}\frac{e^{-x}}xdx]\]the first integral clearly diverges, though the second does not
how you want to apply the comparison test I don't know...

Answer 35

OHH! Ok I think I will use that strategy to explain why I do not use comparison test.

Answer 36

What you said above I meant

Answer 37

there may be a way to use it, but if so it seems trivial....
good luck to any who see how though, back to work for me :)
Good luck!

Answer 38

Thanks

Answer 39

welcome