Can you please evaluate,

My friend says that the answer is ${\pi }^2/12$.
love arun

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.

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 ${\int }_e^X\log (1+x)/x\mathrm{dx}>{\int }_e^{X}1/x\mathrm{dx}=\ln (X)-1\to \infty$ as $X\to \infty$.

Arun - I think your friend got the question slightly wrong, as it is true that

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

The most straightforward way to prove it that I know goes as follows:

$\ln (1+x)/x=1-x/2+x^2/3-x^3/4+...$

Thus, by integrating each term separately, we see that

${\int }_0^1\ln (1+x)/x=1-1/2^2+1/3^2-1/4^2+...=n$

Call this equal to $n$, as it is the number we are trying to find. Also observe that

$n+2(1/2^2+1/4^2+1/6^2+...)=1/1+1/2^2+1/3^2+1/4^2+1/5^2+...$

It is a famous result that $1/1+1/2^2+1/3^2+1/4^2+1/5^2+...={\pi }^2/6$, and I could prove this to you if you'd like.

Also, note that

$1/2^2+1/4^2+1/6^2+...=1/2^2(1/1+1/2^2+1/3^2+1/4^2+1/5^2+...)=(1/4)({\pi }^2/6)$

So

$n+2/4({\pi }^2/6)={\pi }^2/6$

Upon simplifying this, we get

$n={\pi }^2/12$

Remembering our defintion of $n$, this tells us that ${\int }_0^1\ln (1+x)/x\mathrm{dx}={\pi }^2/12$
If anything doesn't make sense, don't hesitate to write again.

Brad

I understood everything BRAD.
That is quite a wonderful solution.
Thanks.

I would like the proof of

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

Now, we can consider this as an infinite polynomial, which is factorable. At the zero's of this function [ $\sin (x)/x$ that is], $x=\pm \pi$, $\pm 2\pi$, $\pm 3\pi$, ..., so we write this as

$1-x^2/3!+x^4/5!-x^6/7!+...=(1-x^2/{\pi }^2)(1-x^2/(2\pi {)}^2)(1-x^2/(3\pi {)}^2)...$

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 $x^2$ is given by

$-(1/\pi +1/(2\pi {)}^2+1/(3\pi {)}^2+...)$

So we may write

$1-x^2/3!+x^4/5!-x^6/7!+...=1-(1/\pi +1/(2\pi {)}^2+1/(3\pi {)}^2+...)x^2+($ some series $)x^4+...$

Now, we may pair up the coefficients of $x^2$ to get

$1/3!=1/6=1/{\pi }^2+1/(2\pi {)}^2+1/(3\pi {)}^2+...$

Now multiply through by ${\pi }^2$, and we obtain

${\pi }^2/6=1/1+1/2^2+1/3^2+1/4^2+...$
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

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

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

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

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

First integrate $\int xy\mathrm{dx}$ and treat $y$ as a constant. We are given

$\int (1/2)x^{2}y\mathrm{dy}$

Now treat $x$ as a constant and we get

$(1/4)x^2{y}^2$
Brad

Ya I got it.

I never thought it was that easy.
thanks BRAD.