Question

how do I find when an integral is convergent or divergent?

Answer 1

really depends on the integral

Answer 2

"1
the limit exists (and is a number), in this case we say that the improper integral is convergent;
2
the limit does not exist or it is infinite, then we say that the improper integral is divergent"

Answer 3

when is an example of the limit not existing?

Answer 4

I think its when it just keeps on going, never actually hits an endpoint, i did this stuff so long ago though, so bear with me.

Answer 5

my problem was ∫1/(xlnx) dx on (e², ∞)

Answer 6

i know the answer is divergent. I just don't know how

Answer 7

first find the "anti -derivative" you will get using a substitution
\[u=\ln(x),du=\frac{1}{x}dx\] that your anti derivative is
\[\ln(\ln(x))\] right?

Answer 8

are you sure?

Answer 9

now this is no problem at
\[e^2\] you get
\[\ln(\ln(e^2))=\ln(2)\] but you are integrating from
\[e^2\] to
\[\infty\] so you have to actually

Answer 10

you would still have ∫1/lnx dx

Answer 11

i am pretty sure. it is
\[\int_{e^2}^{\infty}\frac{1}{x\ln(x)}dx\] if i am reading it correctly

Answer 12

ok lets go slow. is this the right integral
\[\int_{e^2}^{\infty}\frac{1}{x\ln(x)}dx\]?

Answer 13

and if you substitute du for 1/x dx, you have 1/u du

Answer 14

yes

Answer 15

ok so you make
\[u=\ln(x)\]
\[du=\frac{1}{x}dx\] and now you have
\[\int\frac{1}{u}du\] right?

Answer 16

yes

Answer 17

and the "anti derivative of
\[\frac{1}{u}\] is
\[\ln(u)\] and now replacing
\[u=\ln(x)\] you get
\[\ln(\ln(x))\] as your anti - derivative

Answer 18

you can check by differentiation that this is the right thing

Answer 19

so now your job is to evaluate this anti derivative at "b" and e^2, and then take the limit as b goes to infinity

Answer 20

in other words you want
\[\lim_{b\rightarrow \infty} \ln(\ln(b))-\ln(2)\]

Answer 21

ok...

Answer 22

so how can you know if this converges to a number, or diverges to infinity?

Answer 23

so ln(ln(b)) - ln(ln(e²)) ?

Answer 24

it is true that
\[\ln(x)\] grows very very slowly, but it does go to infinity.
right

Answer 25

i just wrote
\[\ln(2)\] instead of
\[\ln(\ln(e^2))\]

Answer 26

yes

Answer 27

i forget how that rule works

Answer 28

ooo because lne cancels?

Answer 29

\[\ln(x)\] is the inverse of
\[e^x\] so
\[\ln(e^x)=x\]

Answer 30

and so
\[\ln(e^2)=2\] and therefore
\[\ln(\ln(e^2))=\ln(2)\]

Answer 31

ok i'm following

Answer 32

that part is not really important anyway, since it is just some number. the question is what is
\[\lim_{b\rightarrow \infty}\ln(\ln(b))\]

Answer 33

hmm... is it infinity?

Answer 34

like divergent? cuz it never meets anything?

Answer 35

and the answer is, it is infinite. because as b goes to infinity, so does
\[\ln(b)\] and as
\[\ln(b)\] goes to infinity so does
\[\ln(\ln(b))\]

Answer 36

in general you will get one that converges if you get something like
\[\frac{1}{b}\] or even
\[\frac{1}{\sqrt{b}}\] something where the variable ends up in the denominator

Answer 37

so this one is divergent? and i have a similar question

Answer 38

yes it is divergent

Answer 39

one questions asks to show that an integral of the form ∫e^(-px) dx on (a, ∞) is convergent if p>0 and divergent if p<0

Answer 40

what is the anti derivative of
\[e^{-px}\]? it is
\[\frac{1}{-p}e^{-px}\] i hope

Answer 41

did you use u sub?

Answer 42

if p > 0 this means
\[-p<0\] so your exponent is negative and therefore you have
\[-\frac{1}{pe^p}\]

Answer 43

should be
\[-\frac{1}{pe^{px}}\]

Answer 44

sorry how did you integrate that?

Answer 45

i guess i used a u-sub but really just did it in my head. you want something with derivative
\[e^{-px}\] it is pretty clear that you have to use
\[-\frac{e^{-px}}{p}\]

Answer 46

the derivative of
\[e^x\] is
\[e^x\] and the derivative of
\[e^{-px}\] is
\[-pe^{-px}\] by the chain rule, so if i want it to work out so that the derivative is
\[e^{-px}\] i have to divide by -p

Answer 47

i got u=e^(-px) and du=-pe^(-px)

Answer 48

not for this one. use
\[u=-px,du=-pdx,-\frac{1}{p}du = dx\]

Answer 49

ok i gotcha!

Answer 50

so now it really boils down to this: does the
\[e^b\] end up in the numerator or does it end up in the denominator?

Answer 51

one sec and i will tell ya

Answer 52

if
\[p>0\] then
\[-p<0\] so your antiderivative using positive exponents will be
\[-\frac{1}{pe^{pb}}\] which goes to zero lickety split as
\[b\rightarrow \infty\]

Answer 53

the numerator?

Answer 54

ahhhhh crap

Answer 55

but on the other hand if
\[p<0\] then
\[-p>0\] so you will have
\[-\frac{e^{pb}}{p}\] these minus signs are annoying i don't know why they put them in the problem.

Answer 56

okay so I don't know where I screwed up. I found the limit as b -> ∞ and got (-e^b + e^a) / p.
was I not supposed to do that?

Answer 57

the idea is simple, they confused it with the minus sign. it would have been simpler to say if
\[p<0\] then
\[\int_0^{\infty} e^{px}dx\] converges and if
\[p>0\] then the integral diverges. less confusing that way

Answer 58

it totally depends on p. you dropped the p in the exponent from your answer

Answer 59

so I have no idea what I was supposed to show in the answer

Answer 60

it all comes down to whether, when you write using positive exponents, the
\[e^{pb}\] is in the numerator or in the denominator. depends totally on the sign of p

Answer 61

if it's positive, it diverges, if it's negative it converges. I grasped that, I just don't know how I show that on paper

Answer 62

you want to say that if
\[p>0\] then \[-p<0\] and so sing positive exponents you get
\[\lim_{b\rightarrow \infty}-\frac{1}{e^{pb}}=0\]

Answer 63

*using

Answer 64

and likewise if
\[p<0\] then
\[-p>0\] so you get
\[\lim_{b\rightarrow \infty}-\frac{e^{pb}}{p}=\infty\]

Answer 65

okay that all makes so much sense now. I understand what you meant by the negative sign in the exponent confusing me... thanks!