The Question
I'm trying to solve this exercise from Folland's real analysis book in chapter 6:
Let $I_\alpha$ be the $\alpha th\ $ fractional integral operator $I_\alpha f\left(x\right)=\frac{1}{\Gamma\left(\alpha\right)}\int_{0}^{x}{\left(x-t\right)^{\alpha-1}f(t)dt}$ for $\alpha>0 $
And let $J_\alpha f\left(x\right)=x^{-\alpha}I_\alpha f\left(x\right)$
a) prove: $J_\alpha$ is bounded on $L^P\left(0,\infty\right)$ for $1<p\le\infty$ more precisely $||J_\alpha f\left(x\right){||}_p\le\frac{\Gamma(1-p^{-1})}{\Gamma(\alpha+1-p^{-1})}||f{||}_p $
b) there exist $f\in L^1(0,\infty)$ such that $J_1f\notin L^1(0,\infty)$
My Understanding
My attempt was to define $$ k(x,y)= \frac{1}{\Gamma\left(\alpha\right)}x^{-\alpha}\left(x-y\right)^{\alpha-1}\chi_{\left\{\left(x,y\right):x<y\right\}}\left(x,y\right) $$ and to use
but when i tried to calculate i get:$$ \int_{0}^{\infty}{\left|k\left(x,1\right)\right|x^{-1/p}dx}=\frac{1}{\mathrm{\Gamma}\left(\alpha\right)}\int_{0}^{1}{x^{-\left(\alpha+1/p\right)}\left(1-x\right)^{\alpha-1}dx} $$
which is a bit different from what i need, what did i do wrong?
Notice that \begin{align} J_\alpha f (x) =&\ \frac{x^{-\alpha}}{\Gamma(\alpha)} \int^x_0(x-y)^{\alpha-1} f(y)\ dy\\ =&\ \int^\infty_0 K(x, y) f(y)\ dy \end{align} where \begin{align} K(x, y) = \frac{1}{\Gamma(\alpha)}x^{-\alpha}(x-y)^{\alpha-1}\chi_{\{(x, y)\in (0, \infty)\times (0, \infty)\mid y<x\}}. \end{align}
Then, we see that \begin{align} \int^\infty_0 |K(x, 1)|x^{-1/p}\ dx =&\ \int^\infty_0\left| \frac{1}{\Gamma(\alpha)}x^{-\alpha}(x-1)^{\alpha-1}\chi_{\{ x\in (0, \infty)\mid 1<x\}}\right| x^{-1/p}\\ =&\ \frac{1}{\Gamma(\alpha)} \int^\infty_1 x^{-\alpha-1/p}(x-1)^{\alpha-1}\ dx\\ =&\ \frac{-1}{\Gamma(\alpha)}\int^\infty_1 \frac{1}{x^{1/p-1}}\left(1-\frac{1}{x}\right)^{\alpha-1}\ d\left(\frac{1}{x}\right)\\ =&\ \frac{1}{\Gamma(\alpha)}\int^1_0 x^{\frac{1}{p}-1}(1-x)^{\alpha-1}\ dx = \frac{1}{\Gamma(\alpha)}\frac{\Gamma(\frac{1}{p})\Gamma(\alpha)}{\Gamma(\frac{1}{p}+\alpha)} \end{align}
However, it seems that my answer is the Holder conjugate of the expected answer. I don't know whether I have made any mistakes or it is a typo in the textbook.