Let $0<\alpha<1/2$, $\lambda\ge1$ and $t\in[0,T]$, where $T>0$ is fixed; I have to prove that \begin{align*} &\frac1{\lambda}\int_0^{\lambda t}e^{-y}[\lambda^{2\alpha}y^{-2\alpha}+\lambda^{\alpha}(\lambda t-y)^{-\alpha}]\,dy\le\\ &\le\lambda^{2\alpha-1}\left[\int_0^{+\infty}e^{-y}y^{-2\alpha}\,dy+\sup_{z>0}\int_0^ze^{-y}(z-y)^{-\alpha}\,dy\right]\,\\ &\le\lambda^{2\alpha-1}\left(\frac1{1-2\alpha}+4\right) \end{align*} I've noted that $\int_0^{+\infty}e^{-y}y^{-2\alpha}\,dy=\Gamma(1-2\alpha)$ but even though here is proved that $\Gamma(1-2\alpha)\le\frac1{1-2\alpha}$, on this other place is explained that not both inequalities $$ \int_0^{\lambda t}(\lambda t-y)^{-\alpha}\,dy\le\sup_{z>0}\int_0^ze^{-y}(z-y)^{-\alpha}\,dy\le4 $$ can be true, in fact $$ \int_0^{\lambda t}(\lambda t-y)^{-\alpha}\,dy=\frac{(\lambda t)^{1-\alpha}}{1-\alpha} $$ and we can keep $\alpha$ and $\lambda$ fixed, and find (with $T$ large enough) a $t$ such that $\frac{(\lambda t)^{1-\alpha}}{1-\alpha}=5 $.
So, how can I find a way out? I found these inequalities on a published paper, so it would be a problem if this is not true.
At least, if not both inequality are true, it would be nice to prove the first one.
