Question

I know you can find this problem online anywhere, but I am just practicing it.

Answer 1

\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx\]we integrate by parts, differentiating the x on the bottom, and integrating e^x.\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}-\large \int\limits_{ }^{ } \frac{e^x}{-x^2}~dx\]\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}+\large \int\limits_{ }^{ } \frac{e^x}{x^2}~dx\]then same integration by parts,\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}+ \frac{e^x}{x^2}~dx-\large \int\limits_{}^{ }\frac{-2e^x}{x^3}dx \]\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}+ \frac{e^x}{x^2}~dx+\large \int\limits_{}^{ }\frac{2e^x}{x^3}dx \]and on,\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}+ \frac{e^x}{x^2}+\frac{2e^x}{x^3} +\large \int\limits_{}^{ }\frac{6e^x}{x^4}dx \]\[\large \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}+ \frac{e^x}{x^2}+\frac{2e^x}{x^3} +\frac{6e^x}{x^4}+\large \int\limits_{}^{ } \frac{24e^x}{x^5}dx \]\[ \int\limits_{ }^{ } \frac{e^x}{x}~dx= \frac{e^x}{x}+ \frac{e^x}{x^2}+\frac{2e^x}{x^3} +\frac{6e^x}{x^4}+ \frac{24e^x}{x^5} +\large \int\limits_{}^{ } \frac{120e^x}{x^6}dx\]\[ \int\limits_{ }^{ } \frac{e^x}{x}~dx= \color{red}{\frac{e^x}{x}+ \frac{e^x}{x^2}+\frac{2e^x}{x^3} +\frac{6e^x}{x^4}+ \frac{24e^x}{x^5} + \frac{120e^x}{x^6}...}\]

Answer 2

I did the integration part, but now I want to think of writing it as a power series.
\[ \int\limits_{ }^{ } \frac{e^x}{x}~dx= \color{red}{\left(\begin{matrix} \\ \frac{e^x}{x}+ \frac{e^x}{x^2}+\frac{2e^x}{x^3} +\frac{6e^x}{x^4}+ \frac{24e^x}{x^5} + \frac{120e^x}{x^6}... \end{matrix}\right)}\]\[ \int\limits_{ }^{ } \frac{e^x}{x}~dx=~ \frac{e^x}{x}\color{red}{\left(\begin{matrix} \\ 1+ \frac{e^x}{x}+\frac{2}{x^3} +\frac{6}{x^2}+ \frac{24}{x^4} + \frac{120}{x^5}... \end{matrix}\right)}\]

Answer 3

if I am doing anything wrong, then please correct me....

Answer 4

I am off at the last line

Answer 5

\[ \int\limits_{ }^{ } \frac{e^x}{x}~dx=~ \frac{e^x}{x}\color{red}{\left(\begin{matrix} \\ 1+ \frac{1}{x}+\frac{2}{x^3} +\frac{6}{x^2}+ \frac{24}{x^4} + \frac{120}{x^5}... \end{matrix}\right)}\]corrected.

Answer 6

I will refersh.

Answer 7

now, I need to do think of a sigma for this whole red thing.
\[\large \sum_{n=0}^{\infty}~\color{red}{ (n+1)x^{-n}}\]

Answer 8

my powers in the red, where I have integral of e^x/x dx= ... are disordered, it should be x^2 on the bottom and then x^3.

but I get the idea, do I ?

Answer 9

\[ \int\limits_{ }^{ } \frac{e^x}{x}~dx=~ \frac{e^x}{x} \times \left(\begin{matrix} \sum_{n=0}^{\infty}~n(n+1)(x^{-n})\\ \end{matrix}\right)\]

Answer 10

error was in the sigma as well. I needed
1 * (1+0) * x^0
2 * (1+1) * x^-1
3 * (1+2) * x^-2

Answer 11

I am done....

Answer 12

if any corrections, besides ones that I have caught myself on.... if not, then over I guess.

Answer 13

the summation you did should be looked at again
doesn't seem to me all correct

Answer 14

i believe you missed signs as well in the integration by part! let me see that

Answer 15

are you sure? can you tell me where I made a mistake please?
it seems as though, that we get:

(1)*(1+0)*x^(-0) --> 1 . when n=0
(1)*(1+1)*x^(-1) --> 2*x^-2 . when n=1
(2)*(1+2)*x^(-2) --> 6*x^-3.. when n=2
(3)*(1+3)*x^(-3) --> 24*x^-4. when n=3

Answer 16

for n=1 spot that you have 2/x not 2/x^2

Answer 17

for the sign i take back seems fine^_^

Answer 18

you mean that for n=1, I get 2/x instead of 1/x maybe?

Answer 19

still a mistake, but it would be better to know where it is

Answer 20

yes

Answer 21

but the integration part is alright?

Answer 22

I'll be typing for the sigma purposes for now...

Answer 23

yes seems good to me!
you just need to rewrite the summation correctly

Answer 24

\[\frac{e^x}{x}\left(\begin{matrix} \\ \end{matrix}1+ \frac{1}{x}+\frac{2}{x^2}+\frac{6}{x^3}+\frac{24}{x^4}+\frac{120}{x^5} \right)\]now once I seee it I will think

Answer 25

yeah, even the n(n+1) part was wrong

Answer 26

I see the pattern so I should be able to figure out the sigma

Answer 27

yes! i believe so :)

Answer 28

\[\large \sum_{n=0}^{\infty } \frac{\rm blank~~for~~now}{x^n}\]

Answer 29

yes! the top should be decrease in power compared to the bottom
so even if you see the pattern you need to see this part as well

Answer 30

ohh, it involves factorial!!!
0!=1
1!=1
2!=2
3!=6
4!=24
5!=120
and on,.....!! omg

Answer 31

yes there is factorial indeeed

Answer 32

i guess your summation becomes obvious now^_^

Answer 33

so, lets do this.
\[\frac{e^x}{x}\left(\begin{matrix} \\ \end{matrix} \sum_{n=-1}^{\infty} \frac{(n+1)!}{x^{n+1}} \right)\]

Answer 34

am I allowed to start from n=-1?

Answer 35

why would you do that?

Answer 36

to have 0! for first term, and 1! for the second term.

Answer 37

from zero is fine

Answer 38

I don't think so

Answer 39

because when n=1, I get 2!/x, and not 1/x

Answer 40

the summation is n!/x^n for n=0 0!/1=1
for n=1 you have 1!/x=1/x
for n=2 you have 2!/x^2=2/x^2

Answer 41

oh true lol, I did n=-1, and added 1 to the power of x on the bottom, and 1 to the (n+1), so just based on that I should have caught myself....

Answer 42

yeah i see your mistake!

Answer 43

\[\large \frac{e^x}{x}\left(\begin{matrix} \\ \end{matrix} \sum_{w=0}^{\infty} \frac{w!}{x^{w}} \right)\]

Answer 44

there it is:)

Answer 45

:) great job pal.
when i first saw you integration by part
i thought it would involve combinations N choose K
but that wasn't the case lol
i saw something similar before that turned like that

Answer 46

yeah, it actually looks neat with the sigma. I feel myself really wise with something like\[\frac{e^x}{x}\left(\begin{matrix} \\ \end{matrix} \sum_{j=0}^{\infty} \frac{(j)!}{x^{j}} \right)\]

Answer 47

thank you !

Answer 48

yeah it those stuff feel amazing when you arrive at such result
welcome pal

Answer 49

:)