# Asymptotic Expansion

1. Oct 25, 2012

### the_kid

1. The problem statement, all variables and given/known data
Calculate an asymptotic expansion to all orders of I(x) as x-->infinity.

I(x)=$\int^{\infty}_{0}$e^{-xt} ln(1+$\sqrt{t})$dt

2. Relevant equations

3. The attempt at a solution
I'm not exactly sure how to begin with this. I've read through the section on my book on asymptotics but am still confused. I think I need to expand the integrand as a Taylor Series.

2. Oct 25, 2012

### Zondrina

Do you remember the Taylor series for ex and ln(1+x)?

3. Oct 25, 2012

### the_kid

Yes.

e$^{x}$=1+x+$\frac{x^{2}}{2!}$+$\frac{x^{3}}{3!}$+$\cdots$

ln(1+x)=x-$\frac{x^{2}}{2}$+$\frac{x^{3}}{3}$-$\cdots$

4. Oct 25, 2012

### Zondrina

Write them as sums for ease, remember though that the series for ln(1+x) does not exist at n=0 so start your series at n=1.

Now how can you re-write your integral so you can integrate term by term?

5. Oct 25, 2012

### the_kid

How do I know where to truncate the Taylor series so that I can multiply them together?

6. Oct 25, 2012

### Zondrina

So :

$e^{-xt} = - \sum_{n=0}^{∞} \frac{xt}{n!}$

What happens when you evaluate the first sum? Can you re-write your sum then :)?

And :

$ln(1 + \sqrt{t}) = \sum_{n=1}^{∞} (-1)^{n+1} \frac{t^{n/2}}{n}$

7. Oct 26, 2012

### the_kid

Hmm, I'm not exactly sure what you mean. Could you clarify?

8. Oct 26, 2012

### Zondrina

What is the first term of the series of e-xt

9. Oct 26, 2012

### the_kid

I believe it is 1...

10. Oct 26, 2012

### Zondrina

Oh a bit of a fumble earlier. Sorry it's 2am.

$e^{-xt} = \sum_{n=0}^{∞} \frac{(-xt)^n}{n!}$

If the first term is 1, then re-write :

$e^{-xt} = 1 + \sum_{n=1}^{∞} \frac{(-xt)^n}{n!}$

11. Oct 26, 2012

### the_kid

No worries!

OK, I've got that. I'm just not sure where to go from there.

12. Oct 26, 2012

### Zondrina

Multiply the two sums of the integral together and then integrate term by term.

13. Oct 26, 2012

### the_kid

OK, simple enough. It looks like this series diverges when x tends to infinity, though? Is that expected?

14. Oct 26, 2012

### the_kid

Thinking about this, it seems to make sense. When x goes to infinity, the e^(-xt) becomes small quickly, and so the integral is dominated by the log term--but integrating that term from 0 to infinity diverges, so it makes sense that the expansion would diverge as x tends to infinity. Right?

15. Oct 26, 2012

### Zondrina

I'm not sure, I didn't actually compute it, but by the looks of it I believe the integral will diverge.

16. Oct 27, 2012

### the_kid

How exactly do I multiply these two together if they have infinite terms?