integral

[nicola de rosa]

1)Prove that $sum_{k=0}^{+infty}1/(2k+1)^2=pi^2/8
2)Compute $int_{0}^{1}(lnx)/(x^2-1)dx$

[luca.barletta]

1) Consider the "triangular wave" function:

$f(x)=|x|={(x,,0<=x<=pi),(-x,,-pi<=x<0):}$

and such as $f(x)=f(x+2kpi), kinZZ$.
Since f(x) is even, we can compute its cosine transformation:

$a_0=2/piint_0^pi f(x)dx = pi$
$a_k = 2/piint_0^pi f(x)cos(kx)dx = {(0,,k=2n , ninNN),(-4/(k^2pi),,k=2n+1 , ninNN):}$

so we have:

$pi/2-4/pisum_(n=0)^(+infty) cos((2n+1)x)/(2n+1)^2$

Note that this result is consistent thank to the Dirichlet's th.
In particular, being $f(x)=x, x in [0,pi]$, we can write:

$x = pi/2-4/pisum_(n=0)^(+infty) cos((2n+1)x)/(2n+1)^2 0<=x<=pi$

and from the position $x=0$ we get

$0=pi/2-4/pisum_(n=0)^(+infty) 1/(2n+1)^2$

that is

$sum_(n=0)^(+infty) 1/(2n+1)^2 = pi^2/8$



2) Of course the following holds:

$int_{0}^{1}(lnx)/(x^2-1)dx=pi^2/8$

I'm just finding an elegant way to put down all the formulas...



Excuse for my english. I've taken some lessons from Biscardi.

[nicola de rosa]

Correct.
Only the integral remains. have you the solution? I wait for you.

[Cheguevilla]

the seguent holds

the following...
"seguent" è biscardiano...

[luca.barletta]

"seguent" è biscardiano...

Only in English, please . "seguent" is biscardian. And now I'm going to edit...

[nicola de rosa]

can nobody solve the integral?

[luca.barletta]

I haven't got a complete solution. Only tell me if I'm on the right way.

$int_0^1 ln(x)/(x^2-1) dx = int_0^1 sum_(n=1)^(+infty) (-1)^(n+1)/n (x-1)^(n-1)/(x+1) dx$

$= sum_(n=1)^(+infty) (-1)^(n+1)/n int_0^1 (x-1)^(n-1)/(x+1) dx$

Now, with the sostitution $x-1=t$:

$sum_(n=1)^(+infty) (-1)^(n+1)/n int_(-1)^0 t^(n-1)/(t+2) dt$

Compute one step of the integral:

$int t^(n-1)/(t+2) dt=t^n/(2n) -1/2int t^n/(t+2) dt -1/(2n)$

Expliciting with respect to the integral on the right side of the equality:

$int_(-1)^0 t^n/(t+2) dt = (-1)^(n+1)/n -2int_(-1)^0 t^(n-1)/(t+2) dt$

Let denote $I(n)$ as:

$I(n) = int_(-1)^0 t^n/(t+2) dt$

We have just found the recursion:

${(I(n) = (-1)^(n+1)/n -2I(n-1)),(I(0)=ln2):}$

We need of $I(n-1)$, and with a little "brain-work" we get:

$I(n-1) = sum_(j=1)^(n-1) (-2)^(n-j+1)/j + (-2)^(n-1)ln2$

And now? Have you followed the same way?

[nicola de rosa]

My way:
$sum_{k=0}^{+infty}x^(2k)=1/(1-x^2)$ $0<=x<1$ (our case)
So
$int_{0}^{1}lnx/(x^2-1)dx=int_{0}^{1}lnx*(-sum_{k=0}^{+infty}x^(2k))dx$
We can exchange two signs , integral and series and so
$int_{0}^{1}lnx/(x^2-1)dx=sum_{k=0}^{+infty}int_{0}^{1}(-x^(2k))*lnxdx$
Now per parti
$int_{0}^{1}(-x^(2k))*lnxdx=[-x^(2k+1)/(2k+1)lnx]_{0}^{1}+int_{0}^{1}x^(2k+1)/(2k+1)*1/xdx$=
$[-x^(2k+1)/(2k+1)lnx]_{0}^{1}+int_{0}^{1}x^(2k)/(2k+1)dx$=
$[-x^(2k+1)/(2k+1)lnx]_{0}^{1}+[x^(2k+1)/((2k+1)^2)]_{0}^{1}$
Now the first part of integral is null $AAk=0,1,2,3..$ for $x=0$ and $x=1$ while the second part $AAk=0,1,2,3..$ is null for $x=0$ and not null for $x=1$ and that is
$int_{0}^{1}(-x^(2k))*lnxdx=1/(2k+1)^2$ $AAk=0,1,2,3..$ and
$int_{0}^{1}lnx/(x^2-1)dx=sum_{k=0}^{+infty}1/(2k+1)^2=pi^2/8$ remembering the first result.