Enunciado do Problema
Os quatro círculos têm o mesmo raio. Determine-o no caso do triângulo medir 1 m².
Problem Statement
The four circles have equal radius. Find it if the size of the triangle is 1 m².
Demostre que para 
Sugestão: utilize a série geométrica de razão 
Prove that for
Hint: use the geometrical series with ratio
ver/see Problema do mês Problem of the month
Prove ou infirme:
Solução de Prof. Paulo Sérgio
Para calcular esta integral, note que
Assim,
Invertendo a ordem de integração e usando a definição de transformada de Laplace, temos:
ou seja,
Outra resolução: fatima
* * *
Prove or disprove: 
Solution by Prof. Paulo Sérgio
To evaluate this integral note that
Hence,
By reversing the order of integration and using the definition of the Laplace transform we get:
therefore,
Other Solver: fatima
Demonstre ou infirme: o integral / Prove or disprove: the integral
é convergente / converges.
25.05.10: corrigida a função integranda /integrand function corrected. Deve ser/Should be
em vez de/instead of
Resolução/Solution: O integral é divergente/The integral diverges:
Prove que/prove that
where/em que 
My solution of POW12 was accepted. [Remark of December 19, 2009: it is only stated that it was "completely or partially proved"]
(Tradução aqui)
“ Problem. Find, with proof, the maximum value of
where
and
is variable. In particular, your answer must be greater than or equal to all values obtained from other choices of
“
Here is the solution I submited.
Solution.
Let 
where 
and
Solving this system of equations we find
and
the latter being a local extremum of 
The function 
On the other hand 
is a maximum of 
Update (Dec., 19,2009): some errors (identified by Rod Carvalho here) corrected.
Let 
and
Let 
Hints: for the first two identities use the formula proved here. As for the last one evaluate the beta function value B
PS. Listed in the Carnival of Mathematics #56. See pingback in the 1st comment.
I’ve just submitted the following solution for the
« Integral of the Week #2
The second Integral Of The Week (IOTW) is rather different from the first in that I am going to give you the evaluation. Your task is to prove it.
But WAIT! Almost every time I have seen this integral evaluated, it has been done by squaring it and converting to polar co-ordinates and that’s the one method of evaluation you can’t use for this particular challenge. I am looking for more ‘interesting’ proofs. Have fun.
Solutions can be posted in the comments section or sent to me by email (obtaining my email address is another puzzle for you to solve) and will be discussed in a future post. Feel free to send your solution in just about any format you like – plain text, uncompiled Latex, PDF, postscript, Mathematica, ODF, even Microsoft Word. When I get around to posting the solutions I will attempt to standardize them (to PDF probably).
By the way – you still have time to submit a solution for the first IOTW.
Then I evaluate 
The second Integral Of The Week equals 
If
![=\displaystyle\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty }](http://s0.wp.com/latex.php?latex=%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty%7Ddy%5Cdfrac%7B1%7D%7B2%5Cleft%28+1%2By%5E%7B2%7D%5Cright%29+%7D%5Cleft%5B+-e%5E%7B-x%5E%7B2%7D%5Cleft%28+1%2By%5E%7B2%7D%5Cright%29+%7D%5Cright%5D+_%7Bx%3D0%7D%5E%7B%5Cinfty+%7D&bg=ffffff&fg=000000&s=0 "=\displaystyle\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty }")
![=\dfrac{1}{2}\left[ \tan^{-1}y\right] _{y=0}^{\infty }](http://s0.wp.com/latex.php?latex=%3D%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5B+%5Ctan%5E%7B-1%7Dy%5Cright%5D+_%7By%3D0%7D%5E%7B%5Cinfty+%7D&bg=ffffff&fg=000000&s=0 "=\dfrac{1}{2}\left[ \tan^{-1}y\right] _{y=0}^{\infty }")
http://calcauxprobteor.wordpress.com/2008/07/24/integral-of-the-week-2-walking-randomly/]
[updated at 22:30 GMT: two typos in integrals corrected and explanation improved]
[update of July 26, 2008: 
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![=\displaystyle\int_{0}^{\infty }\int_{0}^{\infty }xe^{-sx}\left[ \cos \left( s\right) -\cos \left( 3s\right) \right] dxds](http://s0.wp.com/latex.php?latex=%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cint_%7B0%7D%5E%7B%5Cinfty+%7Dxe%5E%7B-sx%7D%5Cleft%5B+%5Ccos+%5Cleft%28+s%5Cright%29+-%5Ccos+%5Cleft%28+3s%5Cright%29+%5Cright%5D+dxds&bg=ffffff&fg=000000&s=0 "=\displaystyle\int_{0}^{\infty }\int_{0}^{\infty }xe^{-sx}\left[ \cos \left( s\right) -\cos \left( 3s\right) \right] dxds")
![I=\displaystyle\int_{0}^{\infty }x\int_{0}^{\infty }\left[ \mathcal{L}\left\{ \cos\left( s\right) \right\} -\mathcal{L}\left\{ \cos \left( 3s\right) \right\}\right] dx](http://s0.wp.com/latex.php?latex=I%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7Dx%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cleft%5B+%5Cmathcal%7BL%7D%5Cleft%5C%7B+%5Ccos%5Cleft%28+s%5Cright%29+%5Cright%5C%7D+-%5Cmathcal%7BL%7D%5Cleft%5C%7B+%5Ccos+%5Cleft%28+3s%5Cright%29+%5Cright%5C%7D%5Cright%5D+dx&bg=ffffff&fg=000000&s=0 "I=\displaystyle\int_{0}^{\infty }x\int_{0}^{\infty }\left[ \mathcal{L}\left\{ \cos\left( s\right) \right\} -\mathcal{L}\left\{ \cos \left( 3s\right) \right\}\right] dx")
![I=\displaystyle\int_{0}^{\infty }\dfrac{9}{9+x^{2}}dx-\displaystyle\int_{0}^{\infty }\dfrac{dx}{1+x^{2}}=\left[ 3\arctan \left( \dfrac{x}{3}\right) -\arctan \left( x\right) \right] _{0}^{\infty }](http://s0.wp.com/latex.php?latex=I%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B9%7D%7B9%2Bx%5E%7B2%7D%7Ddx-%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7Bdx%7D%7B1%2Bx%5E%7B2%7D%7D%3D%5Cleft%5B+3%5Carctan+%5Cleft%28+%5Cdfrac%7Bx%7D%7B3%7D%5Cright%29+-%5Carctan+%5Cleft%28+x%5Cright%29+%5Cright%5D+_%7B0%7D%5E%7B%5Cinfty+%7D&bg=ffffff&fg=000000&s=0 "I=\displaystyle\int_{0}^{\infty }\dfrac{9}{9+x^{2}}dx-\displaystyle\int_{0}^{\infty }\dfrac{dx}{1+x^{2}}=\left[ 3\arctan \left( \dfrac{x}{3}\right) -\arctan \left( x\right) \right] _{0}^{\infty }")
