ò01 log(1+x)
By Arun Iyer on Thursday, June 28, 2001 -
07:16 pm:
Can you please evaluate,
ò0¥ log(1+x)/x.
My friend says that the answer is p2/12.
love arun
By David Loeffler on Thursday, June 28,
2001 - 11:05 pm:
It doesn't exist, I'm afraid. The function doesn't tend to
zero fast enough as n-> infinity, so the integral doesn't have
a finite value.
You may, however, be interested to know that ò0¥ ln(1+x2)/x2 dx=p.
(Integrate by parts.)
By Michael Doré on Thursday, June 28, 2001 - 11:10 pm:
Yes, to see it doesn't exist, just note that
for x > e-1 we have log(1+x) > 1 so log(1+x)/x > 1/x and
òeX log(1+x)/x dx > òeX 1/x dx=ln(X)-1®¥ as X®¥.
By Olof Sisask on Saturday, June 30, 2001
- 12:22 am:
Arun - I think your friend got the question slightly wrong, as
it is true that
ò01 ln(1+x)/x dx=p2/12.
Olof
By Arun Iyer on Saturday, June 30, 2001 -
07:11 pm:
Yupp, I goofed up all right!!
It wasn't my friend's mistake, I heard it incorrectly.
THE INTEGRAL IS FROM 0 TO 1.
Can you give me its proof, Olof or anybody in this case?
love arun
By Brad Rodgers on Saturday, June 30, 2001
- 07:43 pm:
The most straightforward way to prove it that I know goes as
follows:
ln(1+x)=x-x2/2+x3/3-x4/4+...
ln(1+x)/x=1-x/2+x2/3-x3/4+...
Thus, by integrating each term separately, we see that
ò01 ln(1+x)/x=1-1/22+1/32-1/42+...=n
Call this equal to n, as it is the number we are trying to find. Also observe
that
n+2(1/22 +1/42 +1/62 +...)=1/1+1/22 +1/32 +1/42 +1/52 +...
It is a famous result that 1/1+1/22 +1/32 +1/42 +1/52 +... = p2/6, and
I could prove this to you if you'd like.
Also, note that
1/22 +1/42 +1/62 +...=1/22 (1/1+1/22 +1/32 +1/42 +1/52 +...)=(1/4) (p2/6)
So
n+2/4(p2 /6)=p2 /6
Upon simplifying this, we get
n=p2 /12
Remembering our defintion of n, this tells us that
ò01 ln(1+x)/x dx=p2 /12
If anything doesn't make sense, don't hesitate to write
again.
Brad
By Arun Iyer on Sunday, July 01, 2001 -
07:15 pm:
I understood everything BRAD.
That is quite a wonderful solution.
Thanks.
I would like the proof of
1/1+1/22+1/32+1/42+1/52+...=p2/6
Thanks again.
love arun
By Brad Rodgers on Monday, July 02, 2001 -
01:35 am:
Alright, most proofs I know of this are relatively complex and
difficult to understand, so for simplicity's sake I will give you
a proof that is not entirely rigorous.
First of all, note that
sin(x)/x=1-x2 /3!+x4 /5!-x6 /7!+...
Now, we can consider this as an infinite polynomial, which is factorable. At
the zero's of this function [sin(x)/x that is], x=±p, ±2p,
±3p, ..., so we write this as
1-x2/3!+x4/5!-x6/7!+...=(1-x2/p2)(1-x2/(2p)2)(1-x2/(3p)2)...
This is actually known as the Euler sine product and I'm sure it has a rigorous
proof, but the way I've proven it here is not entirely rigorous (i.e. why not
write the factors in another manner), but it's good enough for me. Let's try
to expand this product. Go through and expand the terms once or twice, you
should see that the portion that remains a coeffecient of x2 is given by
-(1/p+1/(2p)2 +1/(3p)2 +...)
So we may write
1-x2 /3!+x4 /5!-x6 /7!+...=1-(1/ p+1/(2p)2 +1/(3p)2 +...)x2 +(
some series )x4 +...
Now, we may pair up the coefficients of x2 to get
1/3!=1/6=1/p2 +1/(2p)2 +1/(3p)2 +...
Now multiply through by p2, and we obtain
p2/6=1/1+1/22 +1/32 +1/42 +...
as required. I think this proof is due to Euler (although he
probably had a better proof for his sine product), but don't
quote me on that. In the archive of old threads, there are some
posts on this series that evaluate it in a more rigorous manner,
but they also use much more complex math.
Brad
By Arun Iyer on Monday, July 02, 2001 -
07:49 pm:
BRAD,
Thanks for the proof. I am quite clear on this question
now.
When you say complex proofs, what mathematics does it
include?
I am just asking this for the sake of expanding my knowledge.If
it is something I am familiar with then it would be really
helpful if I have those proofs too.
love arun
By Brad Rodgers on Wednesday, July 04,
2001 - 01:39 am:
The ones that I have seen other than this involved things like
Jacobian Matrices and Fourier Series. Both these areas are
relatively new to me so rather than have me explain them
exceedingly poorly, here are some links that talk about
them:
Jacobians
Fourier Series: Scroll to the bottom of the page
for results on Fourier Series.
Brad
[Editor: Refer to this
thread for details of the 1/n2 series.]
By Arun Iyer on Wednesday, July 04, 2001 -
06:43 pm:
BRAD,
the sites are really very good. Though I may take time to
understand them since this is the first time I am exposed to
multiple integration.
Can you (or anyone) give me the ideas behind multiple
integration?
love arun
By Brad Rodgers on Thursday, July 05, 2001
- 07:10 pm:
Multiple is integration is just like regular integration done
twice. So for one integral, we let a variable be constant and
integrate with respect to another variable. And then we integrate
with respect to another variable. for example, say we wanted to
find
òòx y dx dy
First integrate òx y dx and treat y as a constant. We are given
ò(1/2)x2 y dy
Now treat x as a constant and we get
(1/4)x2 y2
Brad
By Arun Iyer on Thursday, July 05, 2001 -
07:33 pm:
Ya I got it.
I never thought it was that easy.
thanks BRAD.