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Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)>0$

Here, $T_{n+1}=(1+y)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5 \Rightarrow T_4=(1+t)T_3-t^4=T_3+T_3t-t^4>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

[For, large $n>>0$ as $f_n(0)=\epsilon_n>0$ we expect to find some $N_0$ such that $t^ny(1+y)^t<\epsilon_n$ for all $n>N_0$ and $0\leq y \leq 1$]

Hence, proved. So, it must hold for $-2\leq p \leq 0$

Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)>0$

Here, $T_{n+1}=(1+t)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5 \Rightarrow T_4=(1+t)T_3-t^4=T_3+T_3t-t^4>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

[For, large $n>>0$ as $f_n(0)=\epsilon_n>0$ we expect to find some $N_0$ such that $t^ny(1+y)^t<\epsilon_n$ for all $n>N_0$ and $0\leq y \leq 1$]

Hence, proved. So, it must hold for $-2\leq p \leq 0$

Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)=0$

Here, $T_{n+1}=(1+y)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5$. $T_3=(1+t)T_2-t^3=T_2+T_2t-t^3>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

Hence, proved. So, it must hold for $-2\leq p \leq 0$

Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)>0$

Here, $T_{n+1}=(1+y)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5 \Rightarrow T_4=(1+t)T_3-t^4=T_3+T_3t-t^4>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

[For, large $n>>0$ as $f_n(0)=\epsilon_n>0$ we expect to find some $N_0$ such that $t^ny(1+y)^t<\epsilon_n$ for all $n>N_0$ and $0\leq y \leq 1$]

Hence, proved. So, it must hold for $-2\leq p \leq 0$

Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-1 ;f_n(0)=0$

Here, $T_{n+1}=(1+y)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5$. $T_3=(1+t)T_2-t^3=T_2+T_2-t^3>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=T_n(1+y)^t-1>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

Hence, proved. So, it must hold for $-2\leq p \leq 0$

Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y\leq 1$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.

Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$

Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$

Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=T_n(1+y)^{t+1}-t^{n}y(1+y)^{t}-(1+y) ;f_n(0)=0$

Here, $T_{n+1}=(1+y)T_{n}-t^{n}$.

$T_3=(1+t-t^2)>1$ as $t<0.5$. $T_3=(1+t)T_2-t^3=T_2+T_2t-t^3>1$ and hence, $T_n >1$ $\forall n \in \mathbb N$

For, $n>>0, t^n \rightarrow 0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)(T_n(1+y)^t-1)>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$

Hence, proved. So, it must hold for $-2\leq p \leq 0$