Please check this for me.

1. Nov 19, 2004

DrKareem

I just got out of a differential equations exam. In that exam, i had to integrate e^x/x (e power x over x)

I worked it out on scratch, and i came up with this:

e^x = Sum(i=0->infinity){X^n/n!}

So dividing by x we get Sum(i=0->infinity){x^(n-1)/n!}.
By integrating we get e^x.Sum(i=0->infinity){1/n}.
The latter series diverges, so i concluded that we can't find a solution to that integeral. I re-checked the work that lead to that equation like four times, and it was all correct.

So could it be that the there was a fault in the exam?

HALP!!!
(sorry didn't use latex or whatever it is called to represent my calculations)

2. Nov 19, 2004

matt grime

well, you should check again cos your logic is wrong - the term by term integral doesn't equal what you said it is, and omits to integrate the 1/x term at the beginning as well.

3. Nov 19, 2004

DrKareem

did't understand.
Integral of x^(n-1)/n!=X^n/n.n! right?

sumation(x^n/n!) is e^x no? so you are left with the harmonic series which diverges.

I can't find that logical error i might have made, can you clarify please?

4. Nov 19, 2004

NateTG

Well, for starters you should recogonize that
$$\frac{x}{n} < \frac{x}{n!}$$
so it's not the harmonic series.

I assume you wanted to do something like:
$$\frac{1}{x}e^x=\frac{1}{x} \sum \frac{x^n}{n!}=\sum \frac{x^{n-1}}{n!}=\sum \frac{x^{n-1}}{n(n-1)!}$$
but it's not at all clear to me what that last sum is going to be equal to.

Mathematica/Wolfram indicate that the result is pretty ugly.

Last edited: Nov 19, 2004
5. Nov 19, 2004

DrKareem

Hmm, can you clarify those? didn't get what you're typing...can't see it

6. Nov 19, 2004

DrKareem

OH, and you'd get 1/n, not x/n.

the series 1/n diverges....

7. Nov 19, 2004

DrKareem

$$e^x=\sum\frac{x^n}{n!}$$,

$$\frac{e^x}{x}=\frac{e^n-1}{n!}$$, This should be e^(n-1)/n!, dunn why it isn't working with Latex

$$\int\frac{e^x}{x}dx=\int\frac{e^n}{n.n!}$$,

But $$\frac{x^n}{n!}=e^x$$,

Thus$$\int \frac{e^x}{x}dx=e^x.\sum \frac{1}{n}$$,

But $$\sum \frac{1}{n}$$ would diverge, thus the integral can't be found.

Edit: Hmmm, i read the Latex help file, and i doesn't seem to work :/

Last edited: Nov 19, 2004
8. Nov 19, 2004

Muzza

You need to end things with [ /tex] (no space). You have reversed the /-sign (to \).

9. Nov 19, 2004

DrKareem

yep sorted it out now :) thx

Come to think of it, i do have a hole in my logic lol....how can i integrate with respect to n, can I? dun think so :P

*kareem is the newb*

10. Nov 19, 2004

NateTG

To group in Latex use {}'s - for example e^{n-1}.
As in
$$\frac{e^x}{x}\neq\frac{e^{n-1}}{n!}$$
but
$$\frac{e^x}{x} = \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!}$$

11. Nov 20, 2004

DrKareem

Yes, i was aware of that, but i just didn't bother, that was unproffesional :(