Estimation of the integral

Question

I am trying to compute, or find a good estimate from above the following integral $$ \frac{1}{\pi}\int_{-\infty}^{\infty}|t|^{-1/p}\left|\frac{|t|^{\nu}-1}{t-1}\right|dt, $$where $0<1/p<1$ and $1/p-1<\nu<1/p$.

Thank you.

Answer 1

I have found the following somewhat messy estimates. First define $r=1/p$, so that $-1<r-1<\nu<r<1$. Then notice that the potentially problematic points are $-\infty,0,1$ and $+\infty$. We consider $1$ first:

Let $|t-1|<1/2$. Then by Taylor's theorem and the triangle inequality, for some $\tau$ between $1$ and $t$, $$\left|\frac{|t|^\nu-1}{t-1}\right|\leq|\nu|\left(1+\frac{1-\nu}{2}|\tau-1|\right)<2.$$ Now look at $0$. If $|t|<1/2$ then (simply picture the graphs of the denominator and numerator) $$\left|\frac{|t|^\nu-1}{t-1}\right|\leq\frac{1-|t|^\nu}{1-|t|}<2,$$ and furthermore $$\int_{-1/2}^{1/2}|t|^{-r}\text dt=2^r/(1-r)\leq2/(1-r).$$ Finally, note $-1-r+\nu<-1$ and that the integrand equals $$|t|^{-1-r+\nu}\left|\frac{1-|t|^{-\nu}}{1-t^{-1}}\right|.$$ Consider the fraction in this product. By repeated use of the triangle inequality, if $-1<t<-1/2$ this is bounded by $(2^\nu-1)/2<2$, if $t<-1$ it is bounded by $1$, and if $t>3/2$ it is bounded by $3$. Choose the uniform bound $3$ in all cases. Then $$\int_\mathbb{R}|t|^{-r}\big||t|^\nu-1\big||t-1|^{-1}\text dt\leq3 +\frac{4}{1-r}+6\int_{1/2}^\infty|t|^{-1-r+\nu}\text dt.$$ I'll leave the final steps to you. Is this sufficient? Would you like me to offer more details?

(I should note I'm more of an asymptotic analyst; a hard analyst will almost certainly be able to find sharper estimates than my rough bounds.)