Question

Evaluate \[\int\limits\limits_{}^{}e ^{x}/x\] dx as an infinite series

Answer 1

What math class are you in?

Answer 2

calc 2

Answer 3

e^x 1/x
e^x -1/x^2
e^x 1/x^3
e^x -1/x^4
pattern!
\[\sum_{i=1}^{n}e^x \frac{1}{x^n}(-1)^{n+1}\]

Answer 4

n->infty

Answer 5

use \[e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\]

Answer 6

\[\sum _{n=0}^{\infty } \frac{x^n}{n!} \frac{1}{x}\]

Answer 7

yah i used maclaurin series of e^x i think i have some pluggin error

Answer 8

but integrating factorial might be trouble

Answer 9

it's dx not dn

Answer 10

n! is a constant with respect to x

Answer 11

oh, I see

Answer 12

do you guys have the steps?

Answer 13

\[\sum _{n=0}^{\infty } \frac{1}{n!}x^{n-1}\]

Answer 14

integrate term by term

Answer 15

the answers C + ln lxl + \[\sum_{n=1}^{\infty}x ^{n}/n \times n!\]

Answer 16

im no where close to that

Answer 17

write out the first few terms and then try and integrate..see if that helps you

Answer 18

I would write it as

\[\sum _{n=0}^{\infty } \frac{1}{n!}x^{n-1}=\frac{1}{x}+\sum _{n=1}^{\infty } \frac{1}{n!}x^{n-1}\]

Answer 19

\[\int x^{n-1} dx = \frac{x^n}{n}\]

Answer 20

i have an extra x in the sum

Answer 21

wait i messed up above
\[\int\limits_{}^{}\frac{e^x}{x}dx=\lim_{x \rightarrow \infty} \sum_{i=1}^{n}e^x*\frac{1}{x^n}\]
i can't believe that it was alternating those negatives are being multplied by another negative making it postiive

Answer 22

i wrote it was alternating*

Answer 23

you guys know what im talking about?

Answer 24

integrating by parts

Answer 25

\[\int\sum _{n=0}^{\infty } \frac{1}{n!}x^{n-1}dx=\int\frac{1}{x}dx+\sum _{n=1}^{\infty } \frac{1}{n!}\int x^{n-1}dx\]

\[=\ln|x|+\sum _{n=1}^{\infty } \frac{x^n}{n\times n!}+c\]

Answer 26

sure...but that method won't lead to the solution LameCinnamon is looking for.

Answer 27

ok

Answer 28

i think you want the limit as \[n\to\infty\]

Answer 29

yes we do want that

Answer 30

got it now i integrated wrong

Answer 31

zarkon so the answer i get will be different from the one he will get?
so that means my way is wrong?

Answer 32

no...just in a different form

Answer 33

usually you want to get rid of the exponential function so that you can write the solution as an infinite polynomial

Answer 34

we can being e^x out right?
and we really just have to look to what happens with
\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n}\frac{1}{x^n}\]?

Answer 35

http://en.wikipedia.org/wiki/Exponential_integral

Answer 36

yes...but it doesn't converge to something nice

Answer 37

ok you are right lol

Answer 38

like always

Answer 39

lol

Answer 40

I just noticed that you didn't do parts 100% correctly. your differentiation of the 1/x term's do not have the correct constants

Answer 41

you are right

Answer 42

i was trying to speed against you

Answer 43

lol

Answer 44

zarkon competing with you isn't easy you know?

Answer 45

it is not a competition :)

Answer 46

yes it is :)

Answer 47

minor question for integrals are you able to separate the function if its division

Answer 48

ah...ok ;)

Answer 49

like \[\int\limits_{}^{} x ^{2n}/x\]

Answer 50

\[\int\limits_{}^{} x ^{2n}/x\,dx=\int\limits_{}^{} x ^{2n-1}dx\]

Answer 51

so you cant do this \[\int\limits_{}^{}x ^{2n}\int\limits_{}^{}1/x\]

Answer 52

\[\int\frac{f(x)}{g(x)}dx\neq\int f(x)dx\int\frac{1}{g(x)}dx\]

if that is what you are getting at.

Answer 53

no you can't do that

Answer 54

\[\int\limits\limits_{}^{} \frac{x ^{2n}}{x} dx=\int\limits_{}^{}x^{2n-1}dx=\frac{x^{2n-1+1}}{2n-1+1}+C, 2n-1\neq-1\]

Answer 55

ok yah i keep considering that

Answer 56

it doesn't work for differentiation and it doesn't work for integration

Answer 57

\[\int\limits\limits\limits_{}^{} \frac{x ^{2n}}{x} dx=\int\limits\limits_{}^{}x^{2n-1}dx=\frac{1}{x}+C, 2n-1=-1\]

Answer 58

hmm...

Answer 59

hmmm...

Answer 60

\[\ln|x|+c\]

Answer 61

darn it i know that

Answer 62

i know you know ;)

Answer 63

going too fast :)

Answer 64

2n-1=-1
\[\int\limits_{}^{}\frac{1}{x} dx=\ln|x|+C\]

Answer 65

booya!

Answer 66

lol

Answer 67

shower time! :)

Answer 68

later

Answer 69

lol write some integrals using some shampoo

Answer 70

i'll try that