A very nice integral

[fu^2]

solve $int_(-pi)^(pi)sin((1/2+n)t)/(2sin(t/2))dt$ with $n\in\NN$.

[ReA1]

This is completeley insane

[michele.c.-votailprof]

"fu^2":solve $int_(-pi)^(pi)sin((1/2+n)t)/(2sin(t/2))dt$ with $n\in\NN$.

Oh my God...It's the first time I see that!!! It's something of soprannatural...

[elgiovo]

"fu^2":very good elgivio, this is beautiful proof!

Thank you. But now, for God's sake, change the title of the topic in "A very nice integral"!!

elgivio

[fu^2]

very good elgivio, this is beautiful proof!


ps in the first passage there is an error in the sum, but i think that it is a "svista"

now the integral is very simple !

[elgiovo]

$1/2+sum_(k=1)^ncos(kt)=-1/2+sum_(k=0)^n cos(kt)=-1/2+Re[sum_(k=0)^n e^(ikt)]=-1/2+Re[(e^(it(n+1))-1)/(e^(it)-1)]$

$=-1/2+Re[(e^(i (n+1)/2 t))/(e^(i t/2)) (e^(i (n+1)/2 t)-e^(-i (n+1)/2 t))/(e^(i t/2)-e^(-i t/2))]=-1/2+(sin((n+1)/2 t))/(sin(t/2)) Re[(e^(i (n+1)/2 t))/(e^(i t/2))]$

$=-1/2+(sin((n+1)/2 t) cos(n/2 t))/(sin(t/2))=(2sin(n/2 t + t/2)cos(n/2 t) - sin(t/2))/(2 sin(t/2))=(sin[(n+1/2)t])/(2sin(t/2))$.

[fu^2]

yes, the result is ok! but i don't know the residue's theory...

In fact exist a trick which you can resolve this integral in one line using the knowledge of analysis one...

proof this lemma

$1/2+sum_(k=1)^(n)cos(kt)=sin((1/2+n)t)/(2sin(t/2))$

[Kroldar]

The result for $n=0$ is trivial, so let's consider $n > 0$.

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

To semplify the expression, let's assume $2n+1=k$ and $t/2=u$, so we obtain

$int_(-pi)^(pi)sin((2n+1)/2t)/(2sin(t/2))dt = int_(-pi/2)^(pi/2)sin(ku)/(sin(u))du$

Now we can verify that $u=pi/2$ is a point of simmetry for the function $f(u)=sin(ku)/(sin(u))$, in fact $f(pi/2-u)=f(pi/2+u)$, so

$int_(-pi/2)^(pi/2)sin(ku)/(sin(u))du = 1/2 int_(-pi/2)^(3pi/2)sin(ku)/(sin(u))du$

and if we note that $f(u)$ is periodical of $2pi$, we can conclude that

$1/2 int_(-pi/2)^(3pi/2)sin(ku)/(sin(u))du = 1/2 int_(0)^(2pi)sin(ku)/(sin(u))du$

Let's assume $z=e^(ju)$, so the integral becomes

$1/2 int_(0)^(2pi)sin(ku)/(sin(u))du = 1/(2j) int_(|gamma|=1) (z^k-1/z^k)/(z-1/z) dz/z = 1/(2j) int_(|gamma|=1) ((z^(2k)-1)/z^k)/(z^2-1) dz = 1/(2j) int_(|gamma|=1) (z^(2k)-1)/((z^k)(z^2-1)) dz$

We can now conclude that

$1/(2j) int_(|gamma|=1) (z^(2k)-1)/((z^k)(z^2-1)) dz = 1/(2j) (pijR[1]+pijR[-1]+2pijR[0])$

where $R[u_0]$ is the residue in $u_0$.

But

$R[1] = R[-1] = 0$

so

$1/(2j) int_(|gamma|=1) (z^(2k)-1)/((z^k)(z^2-1)) dz = piR[0]$

The point $0$ is a pole of order $k$ and we have

$R[0] = 1/((k-1)!) lim_(zto0) [d^(k-1)/dz^(k-1) 1/(z^2-1)]$

The final result should be

$int_(-pi)^(pi)sin((1/2+n)t)/(2sin(t/2))dt = pi 1/(2n!) lim_(zto0) [d^(2n)/dz^(2n) 1/(z^2-1)] = pi$