Csdrhrt

Does anyone know how to show $int_{0}^{infty} u^{-alpha} sin(u) , du >0$ for $0<alpha<1$ (without explicitly having to calculate the exact value)?

For another way, integrate by parts:
begin{align*}I(alpha,M) &= int_0^M u^{-alpha}sin u,du\
&= left[u^{-alpha}cdot (1-cos u)right]_{u=0}^{u=M} + int_0^M alpha u^{-alpha-1}cdot (1-cos u),du\
I(alpha,M) &= M^{-u}cdot (1-cos M) + alphaint_0^M u^{-alpha-1}cdot (1-cos u),duend{align*}
For positive $alpha$ and $M$, both terms are clearly nonnegative; the first is the product of two nonnegative terms, and the second is the integral of a nonnegative function. Evaluating that first term to be zero at zero - well, actually, that's a limit, and it requires $u<2$ so that the zero of the $1-cos$ overwhelms the growth of the $u^{-alpha}$.

As $Mtoinfty$, this calculation doubles as a proof that this improper integral converges. For $0<alphale 1$, we've converted a conditionally convergent integral into an absolutely convergent one with this manipulation; the new integral converges as $Mtoinfty$ by comparison with $int 2u^{-alpha -1}$ and the boundary term goes to zero.

I'd also like to call out two choices that may seem odd at first but turn out to be broadly useful techniques:

- "Integrate the part that wiggles". A lot of the time, when we integrate by parts, we're not actually trying to find an antiderivative. With the integral of an oscillating function times a decaying function, one common option is to integrate by parts to improve convergence behavior, integrating the oscillating part and differentiating whatever's left over. If we have to introduce some factor here to make the oscillating part's integral work, we do it.

- Choosing $1-cos$ as the antiderivative instead of $cos$. That $+C$ in the indefinite integral gets ignored a lot, but it's a free choice; we should use whatever works for us, not just whatever looks like zero. Here, since we care about the integral being positive, we choose $1-cos$ to make that positive after integration, and also to zero out the boundary term at zero. Actually, for the range of $alpha$ we care about, any other choice of antiderivative would have split the integral as the sum of a divergent limit and a divergent (at zero) improper integral. That's definitely not a desirable outcome.

Since the Laplace transform is a self-adjoint operator, the Laplace transform of $sin(x)$ is $frac{1}{s^2+1}$ and the inverse Laplace transform of $u^{-alpha}$ is $frac{s^{alpha-1}}{Gamma(alpha)}$,
$$ int_{0}^{+infty} u^{-alpha}sin(u),du = frac{1}{Gamma(alpha)}int_{0}^{+infty}frac{s^{alpha-1}}{s^2+1},ds=cosleft(frac{pialpha}{2}right)Gamma(1-alpha)$$
for any $alphain(0,1)$. On the other hand the finite-ness of the LHS just follows from integration by parts:
$$ int_{0}^{+infty} u^{-alpha}sin(u),du = alphaint_{0}^{+infty}frac{1-cos u}{u^{alpha+1}},du$$
since $1-cos u$ behaves like $frac{u^2}{2}$ in a right neighbourhood of the origin and it is non-negative and bounded far from the origin.

$$ int_0^infty u^{-alpha} sin(u); du = sum_{n=0}^infty
int_0^{pi} left((2n pi + t)^{-alpha} - ((2n+1)pi + t)^{-alpha}right) sin(t); dt $$

and the integrand is positive.