# Sum an infinite series by definite integrals

1. Apr 20, 2004

### KLscilevothma

In the first part of the question, I proved that

$$\int_{1/2}^{2} \frac{ln x}{1+x^2} dx = 0$$

Then I needed to evaluate the following but I didn't know how to do it. Can you give me some clues? I know it must be related to the definite integral that I proved in the first part, but how?

$$\lim_{n\rightarrow\infty} \sum_{k=1}^{3n} \frac{ln (2( \frac{1}{2} + \frac{k}{2n}))}{2n (1 + ( \frac{1}{2} + \frac{k}{2n})^2)}$$

2. Apr 21, 2004

### ReyChiquito

What you need to do is use the following criteria...

A non-negative decreasing series $$\sum_{k=1}^{\infty}x_{k}$$ converges if and only if $$\int_{1}^{\infty}f(x)dx$$ converges, where $$x_{k}=f(k)$$.
If you do a change of variable in the series and use this criteria, youll get something quite similar to the integral above...

3. Apr 22, 2004

### KLscilevothma

mm.... though I don't think I understand what you said as I just finished the A-level, but is
$$\int_{1/2}^{2} \frac{ln2 x}{1+x^2} dx = 0$$
correct ?

4. Apr 26, 2004

### ReyChiquito

to tell you the truth, i cant solve the integral. How did you did it?

5. Apr 27, 2004

### KLscilevothma

No, I don't know how to do it. It's merely a wild guess and it's what I wrote in the exam paper few days ago. I just wanted to know if I was guessing correctly and that's why I posted here. I know a person who seems to know how to solve it. If I get the solution, I'll post it here.

6. Apr 29, 2004

### trancefishy

i got a decimal approximation (using 40 trapezoids) of about .44. so zero doesn't seem it. and how can you write a paper about that on a guess? did you have no proof or what?

7. Apr 30, 2004

### ReyChiquito

it cant be zero, just look at the graph and the domain... the integrand is always positive in the domain of integration...

8. Apr 30, 2004

### matt grime

it can be zero, log(x) is negative for x<1.

to prove it properly, split the integral into two bits from 1/2 to 1 and 1 to 2. then do a substitution x to 1/x in the part going from 1/2 to 1, and voila it will cancel with the other integral.

the second sum now must be zero since it is a numerical approximation for the integral and the limit is as the width of the strips tends to zero, hence as the function is Riemann integrable, it must be the area, ie 0.

Last edited: Apr 30, 2004
9. Apr 30, 2004

### HallsofIvy

But the integral is of ln(2x), from x= 1/2 to 2 so 2x is from 1 to 4. The integrand is positive on the interval.

10. Apr 30, 2004

### matt grime

In the initial post it's not 2x inside the log.

And if that sum isn't exactly the numerical integral fo even sudivisions, then it is close to it:

take out the 2 from the log, and you're left with evaluating the integral of log2/{1+x^2} + the thing you first thought of which is zero.

Last edited: Apr 30, 2004
11. Apr 30, 2004

### KLscilevothma

Argh! I'm soo sorry to confuse you. Actually I wanted to ask if it's right to treat
$$\lim_{n\rightarrow\infty} \sum_{k=1}^{3n} \frac{ln (2( \frac{1}{2} + \frac{k}{2n}))}{2n (1 + ( \frac{1}{2} + \frac{k}{2n})^2)}$$
as
$$\int_{1/2}^{2} \frac{ln2 x}{1+x^2} dx$$ ,
but not if$$\int_{1/2}^{2} \frac{ln2 x}{1+x^2} dx = 0$$.

Trancefishy said, "i got a decimal approximation (using 40 trapezoids) of about .44." and
$$\int_{1/2}^{2} \frac{ln2 x}{1+x^2} dx$$
gives rise to 0.446. It seems that my consideration was correct.

12. Apr 30, 2004

### ReyChiquito

here is what i understand of your q.

by the criteria i posted above and doing a change of variable

$$\lim_{n\rightarrow\infty} \sum_{k=1}^{3n} \frac{ln (2( \frac{1}{2} + \frac{k}{2n}))}{2n (1 + ( \frac{1}{2} + \frac{k}{2n})^2)} < \lim_{n\rightarrow\infty} \int_{1}^{\infty} \frac{ln2 x}{1+x^2} dx$$

you need to prove that
$$\int_{1}^{\infty} \frac{ln2 x}{1+x^2} dx < M$$
im not sure if this is true...

if it is... that means that the series converge uniformly for all n, wich means the limit exist.

now you can exchange the limit and the sum and solve the problem.

ps...
$$\int_{1/2}^{2} \frac{ln2 x}{1+x^2} dx$$ is hard integral to solve... i think only numerical methods or complex variable will do the trick.

Last edited: Apr 30, 2004
13. Apr 30, 2004

### matt grime

the last integral is easy to solve! as you'd realise if you'd read my posts. split it by splitting the log. one becomes an inverse tan integral the other is zero, i think, as you had to show, and as i think my proof demonstrates. ray, i think you've not read it correctly. the sum, in the limit IS equal to the integral as the integrand is a continuos function and the it is equal to that numerical approximation in the limit.

14. Apr 30, 2004

### ReyChiquito

its not clear for me that the other part of the integral is zero, is this a hunch or you really know it? i know that that part has a negative portion of area, but still, as i cannot solve it, im not willing to accept that it is zero...
i know, you split the integral from 1/2 to 1 and from 1 to 2, but then what?
On the other matter, do you mean that the limit of the sum is the Riemann sum in that interval?... thats the only way i can see that the limit of the series is equal to the integral...

Last edited: Apr 30, 2004
15. May 1, 2004

### matt grime

split the integral by splitting the logs, then do the bit with log x in as i suggested above by subs x to 1/x, and i think it will cancel - I am presuming that the integral is zero from the first post, and this certainly appears to do the trick

In the limit that sum is the integral, since the function is continuous a compact subset and the n'th term in the sequence is the approximation using n evenly sized intervals, hence it will converge to the Riemann integral.

16. May 1, 2004

### ReyChiquito

At some point i thought it was the Riemann Sum, but i had some troubles seeing it (i took my calc courses long ago), but as you say... it is
heh... as u say i missinterpreted the whole question, but it gave me a good time trying to solve it, and i dare anyone to find the antiderivative of that ugly function (even if it is well behaved)
good thing teachers know what they are doing and give nice, canceling each other intervals