It's more difficult !

[Camillo]

Prove that the function defined by :

$ u(x)= (tan^(+)(pix))^(1/2)*e^(-x) $

is integrable in $(0, +oo)$.


$tan^(+)(pix) = tan(pix) $ if $ tan(pix) >0 $ ; else $ = 0 $

[Camillo]

I don't know of different paths , may be they exist.

[Eredir]

"Camillo":Nice proof !


Of course the index in the sum is $ n $ instead of $ i$.

Of course, I will correct that immediately.
Is your solution similar? If you followed a different path I would like to know which one.

[Camillo]

Nice proof !


Of course the index in the sum is $ n $ instead of $ i$.

[Eredir]

"Camillo":Prove that the function defined by :

$ u(x)= (tan^(+)(pix))^(1/2)*e^(-x) $

is integrable in $(0, +oo)$.


$tan^(+)(pix) = tan(pix) $ if $ tan(pix) >0 $ ; else $ = 0 $

We observe that $tan(\pix)>0$ if $x\in(n,(2n+1)/2)$ for $n\in\NN$, and is zero elsewhere.
It is possible to prove that $lim_(b->(\pi/2)^-)\int_0^btan^n(x)dx$ converges if $0<=n<1$, so $u(x)$ is integrable in $(0,a)$ with $a\in\RR$ and we only need to show that it converges in $(0,+\infty)$.
We use Hölder's inequality with $p=3/2$ and $q=3$ to get $\int_n^((2n+1)/2)sqrt(tan^(+)(pix))*e^(-x)dx <= (\int_n^((2n+1)/2)(tan^(+)(pix))^(3/4)dx)^(2/3) * (\int_n^((2n+1)/2)e^(-3x)dx)^(1/3) <= C(1/3 e^(-3(n+1/2))(e^(3/2)-1))^(1/3) <= K e^(-(n+1/2))$.
We evaluate $\int_0^(+\infty)sqrt(tan^(+)(pix))*e^(-x)dx = \sum_(n=0)^(\infty)\int_n^((2n+1)/2)sqrt(tan^(+)(pix))*e^(-x)dx <= \sum_(n=0)^(\infty)K e^(-(n+1/2)) = K(sqrt(e))/(e-1)$ and so $u(x)$ is integrable in $(0,+\infty)$.