Question

legit answer or am i way off base: Show that integral from 1 to infinity: (e^-x) / x dx converges.

substitution gets you 0 / infinity L'hospital get you -e^-x/1 or converges to zero???

Answer 1

\[\int\limits_{1}^{\infty}((e^-x)/x) dx\]

Answer 2

Lhopital is for derivatives ...

Answer 3

does the function -1/x^2 converge?

Answer 4

yes by p test 2>1 and p>1 converges

Answer 5

so lets integrate it; and we get the function 1/x

does this converge?

Answer 6

the convergence of a derivative is not a gaurentee that the integral will converge ...

Answer 7

now p = 1 and by p test p=1 diverges...sorry slow wrote p test down wrong in class and was doing it backwards for three days

Answer 8

so looks like i should integrate and see what that gets me...

Answer 9

if we integrate this by parts; it might produce a series .....

Answer 10

this will be slow if you want to help elsewhere...

Answer 11

lol, its 130am .... im not really wanting to help anywhere :)

Answer 12

so that looks like (x)(-e^(-x)) - e^(-x)

Answer 13

\[\frac{1}{x}e^{-x}=\frac{1}{x}\sum_{n=0}^{inf}\frac{(-1)^{n+1}}{n!}x^n\]
\[\frac{1}{x}e^{-x}=\sum_{n=0}^{inf}\frac{(-1)^{n+1}}{n!}x^{n-1}\]

Answer 14

hmm, -1^n that is ... i seem to have generated a spurious negative :)

Answer 15

if I got that then i would be thinking ratio test? and spurious sign changes is my signature move

Answer 16

this is still not integrated; i know when we derive a summation we add to n each time; so i assume when we integrate we get a n-1 to the summantion

since n needs to start at 0; lets pull out the first iteration before we integrate

Answer 17

maybe that is what my key suggests...it just integrates the e^-x and the 1/x disappears in an unexplained poof of smoke

Answer 18

\[\int \frac{1}{x}+\sum_{n=1}^{inf}\frac{(-1)^{n}}{n!}x^{n-1}\ dx\]

Answer 19

\[ln(x) + \sum_{n=0}^{inf}\frac{(-1)^n}{n!(n)}x^n+C\]

Answer 20

uh oh, we got a zero in the denominator with that one :)

Answer 21

it will be ok if we start with 1

Answer 22

yeah, but that ln(x) on the outside doesnt converge ...

Answer 23

so i have a feeling this thing diverges. we could chk it with the wolf

Answer 24

i think they disappeared the 1/x with the idea that if integrate e^-x and converge the multiply 1/x would just make it smaller?

Answer 25

its possible i could have tried to modify it in another manner; i ahvent done enough of these to be too confident

Answer 26

All this converge diverge stuff seems like cheating ....ignore that part just approximate...im just not sure when to do it

Answer 27

well at least in know not to check the limits with l'hospital....seemed like such a good gambit too

Answer 28

:) the summation we got converges by a ratio test ....

if we turn the ln(x) into a summation (power series) and add the 2 together I wonder if it pans out

Answer 29

im saving the series expansion for tomorrow...(part three of the final) I'll work the ratio test on that sum/series for practice thanks for the helps

Answer 30

\[
\int_1^c \frac {e^{-x}}x dx
\\
u=\frac {1}x, \, dv =e^{-x} dx\\
du =-\frac 1 {x^2},\, v =- e^{-x}\\
\int_1^c \frac {e^{-x}}x dx = \left [- \frac {e^{-x}}x \right]_1^c -
\int_1^c \frac {e^{-x}}{x^2}dx=\\
- \frac {e^{-c}}c+ \frac {e^{-1}}1 -\int_1^c \frac {e^{-x}}{x^2}dx
\]

Answer 31

Notice that
\[
\int_1^c \frac {e^{-x}} {x^2} dx \le \int_1^c \frac {dx} {x^2} = 1 - \frac 1 c\\
0\le \int_1^\infty \frac {e^{-x}} {x^2} dx \le 1
\]
Hence
\[
\int_1^\infty \frac {e^{-x}} {x^2} dx
\]
is convergent.
By this and my previous post
the original integral is convergent.

Answer 32

so that uses squeeze to prove convergence?

Answer 33

Yes

Answer 34

If c goes to infinity
\[
\int_1^\infty \frac {e^{-x}}x dx=
\\

-0+ \frac {e^{-1}}1 - k

\]
Where
\[
K = \int_1^\infty \frac {e^{-x}}{x^2}dx
\]

Answer 35

K=k

Answer 36

oK

Answer 37

Did you understand it?

Answer 38

i dont think we did anything like that in class - you are subtracting the other integral from the original?

Answer 39

oh ok that is from the integration by parts...now that makes sense

excellent thanks

Answer 40

yw