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I want to find an upper bound of the function $h(t)$ on $t \in [0, \infty [$ with $d >1$ which is defined by $$ h(t) := \int_0^t (1+t)^d (1+t-r)^{-d} (1+r)^{-d} dr$$ and I could easily prove that $h(t) \leq \frac{2^{d+1}}{d-1}$ by the following way: \begin{eqnarray*} h(t) &=& \int_0^{t/2} (1+t)^d (1+t-r)^{-d} (1+r)^{-d} dr + \int_{t/2}^t (1+t)^d (1+t-r)^{-d} (1+r)^{-d} dr \\ &\leq& \int_0^{t/2} (1+t)^d (1+t-\frac{t}{2} )^{-d} (1+r)^{-d} dr + \int_{t/2}^t (1+t)^d (1+t-r)^{-d} (1+ \frac{t}{2} )^{-d} dr \\ &\leq& 2^d \left( \int_0^{t/2} (1+r)^{-d} dr + \int_{t/2}^t (1+t-r)^{-d} dr \right) \; (\because (1+t)^d (1+t/2)^{-d} \leq 2^d ) \\ &= & 2^{d+1} \int_0^{t/2} (1+r)^{-d} dr \\ &=& \frac{2^{d+1}}{-d+1} ((1+t/2)^{-d+1} -1 ) \\ & \leq & \frac{2^{d+1}}{d-1} \; (\because (1+t/2)^{-d+1} \leq 1 ). \end{eqnarray*} But I think this bound is too big, so is there sharper one for $h(t)$ which is not depend on $t$ ?

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up vote 3 down vote accepted

First of all, the integrand is symmetric with respect to $r=t/2$, so that $$ h(t)=2(1+t)^d\int_0^{t/2} (1+t-r)^{-d} (1+r)^{-d}\,dr. $$ The function $(1+t-r)(1+r)$ is increasing and concave on the interval $[0,t/2]$. It follows that $$ (1+t-r)(1+r)\ge1+t+\frac t2\,r,\quad 0\le r\le t/2. $$ Then $$ h(t)\le2(1+t)^d\int_0^{t/2} \Bigl(1+t+\frac t2\,r\Bigr)^{-d}\,dr. $$ It follows immediately that $h(t)\le t$ for all $t\ge0$. On he other hand, evaluating the integral we get $$ h(t)\le\frac{4(1+t)^d}{(d-1)t}\Bigl((1+t)^{-d+1}-(1+t+\frac{t^2}{4})^{-d+1}\Bigr)\le\frac{4(1+t)}{(d-1)t}. $$ Thus $$ h(t)\le\max\Bigl(t,\frac{4(1+t)}{(d-1)t}\Bigr),\quad t\ge0. $$ Evaluating at the point in which $$ t=\frac{4(1+t)}{(d-1)t} $$ we finally get $$ h(t)\le\frac{2(1+\sqrt d)}{d-1},\quad t\ge0. $$

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Thank you very much! –  Leun Kim Dec 5 '12 at 13:30
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