# 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 :(