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![$$\int\limits_{0}^{\infty}\frac{x^{a-\frac{1}{2}}}{(1-x)^a(1+x\tan^2{\theta})^a}dx=\frac{2\Gamma\left(a+\frac{1}{2}\right)\Gamma(1-a)}{\sqrt{\pi}}\cos^{2a}{\theta}\frac{\sin{\left[(2a-1)\theta\right]}}{(2a-1)\sin{\theta}}$$](http://latex.artofproblemsolving.com/5/3/6/53683a911a95d999f3dcb73ecd16b9947e08551b.png)
for
The source should be from Legendre, but I cannot find it.
and 
be two points that are diametrically opposite to each other on a unit sphere. 
right square pyramids are fitted along the line segment 
, such that the apex and altitude of each pyramid 
, where 
, are 
and 
respectively, and the points 
are collinear. pyramids, with altitudes of equal length, that can be fitted in the sphere, in terms of . pyramids that can be fitted in the sphere, in terms of . tends to infinity.
is even. The corresponding coefficients of the polynomial 
, that are equally distant from left and right, can not be both non zero. So it seems a natural thing(i did it right away) to divide the equation by 
. Denoting 
and 
, our equality transforms like:![\[\sum_{i\in I_1} c_i z^i = \sum_{i\in I_2} c_i z^{-i}\]](http://latex.artofproblemsolving.com/e/7/b/e7b566661d7f7740aa0afe533d6eb4c5cc9ae5b9.png)
where 
and 
.-th roots of unity, so we set 
and get:![\[(1)\,\,\,\,\,\,\,\, \sum_{\ell\in I_1} c_{\ell} e^{i\ell t} = \sum_{\ell\in I_2} c_{\ell} e^{-i\ell t}\]](http://latex.artofproblemsolving.com/3/c/6/3c6d256dcf619fb2e4458677d1a5f27bb45bc3b0.png)
's among 
that don't satisfy 
. The fact that the set of indecies 
and 
are disjoint should play a crucial point, since otherwise in general case it's not true. That's, the key question is how to use the fact and are disjoint. were some kind of identity, we could multiply both sides by the conjugate of the RHS and integrate over ![$[0,2\pi]$](http://latex.artofproblemsolving.com/1/6/6/166f2f81461a7a6ce563dbb02f64f78f6bd08acc.png)
and using the fact 
are orthogonal on and get a contradiction, since the LHS would be 
exactly due to the fact 
. Since this approach strongly uses the fact 
are disjoint it is a very good indicator. are orthogonal also in a discrete sense, namely:![\[(2)\,\,\,\,\,\,\,\,\sum_{j=0}^{n-1} e^{i\ell t_j} e^{ir t_j}= \begin{cases}
0\,,\, \ell+r \not\equiv 0\pmod{n}\\
n\,,\,\ell+r \equiv 0\pmod{n}
\end{cases}
\]](http://latex.artofproblemsolving.com/0/8/c/08cf3ef2be5d16269fac1e8ca0b8080717855f0d.png)
It right away yields all 
cannot be roots of . (which btw is trivial since that equality is of degree 
). Indeed assume on the contrary:![\[(3)\,\,\,\,\,\,\,\, \sum_{\ell\in I_1} c_{\ell} e^{i\ell t_j} = \sum_{\ell\in I_2} c_{\ell} e^{-i\ell t_j}\,,\,j=0,1,\dots,n-1\]](http://latex.artofproblemsolving.com/f/c/7/fc786a64883e07f03fc89e0c796be27ae503ad2f.png)
Multiplying by conjugate of the 
and summing over 
, using 
,we get: 
. (because the case all 
are zeroes is trivial, since the has only 
solutions). to breach 
?. Assuming so, we can rewrite and then again with some offset 
with respect to 
, where 
, that's:
either the LHS of 
or the LHS of 
is zero, we can multiply them, taking first the conjugate of LHS of and sum over 
and due to orthogonality again to get a contradiction like 
.
, there exists such that neither is a solution of nor 
is solution of . That cracks the problem, since then for the set 
of non-solutions of we can say that every offset belongs to 
, which can hold only when 
is not too small. The rest is a routine.
and see later where it comes in.
in a useful way.
is the coefficient of 
in 
. So we conclude that this coefficient equals 
(with the understanding that all the indices and the exponents are to be read 
and 
).
being one which by the root-of-unity filter means that 
.
and therefore on them being roots. In particular, we obtain the easy observation that not all roots of unity are also roots of 
.
as a polynomial identity (to be very precise: as an identity in ![$\mathbb{C}[x]/(x^n-1)$](http://latex.artofproblemsolving.com/c/a/b/cabb549a25f70177e657c108442593371f2c232f.png)
.) in 
i.e. the coefficient of 
in 
and as above, we then find![\[\sum_{\omega} \omega^2f(\omega)f(x\omega)=nx^{n-1}\]](http://latex.artofproblemsolving.com/c/f/f/cffb79e2d90fa00e2e5707c517f80eb05219053e.png)
as an identity in and in particular for 
being a root of unity. The first thing we tried above was exactly the case 
. vanishes. there is one 
such that both and 
are not roots of .
and the exact estimate then follows from an easy combinatorial double-counting argument where the condition 
turns out to be exactly the estimate we need.
be the set of all roots of 
.
.
is with all coefficients conjugated.
there exists 
such that
and
, there exists 
such that 
are not roots of and 
roots of are not roots of . Hence
is sufficient. Also it's a better bound than the original one.
prove that 
Find 
.
and 
,
be a polynomial with complex coefficients such that
Prove that 
and 
common roots.