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Let $s,\sigma,t\in\mathbb{R}$, with $s<\sigma

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    Do you know the Beta function? – 2017-01-06

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The following does not work (there is a reverse inequality), but maybe can be useful? (the answer is in the comments below)

Notice that $$ \frac{\Gamma(2N+4)}{\Gamma^2(N+2)} = \frac{(2N+3)!}{(N+1)!^2} = (2N+3) {2N+2 \choose N+1} \ge \sum_{k=0}^{2N+2} {2N+2 \choose k} = 2^{2N+2}. $$ Also, by the substitution $\sigma = (t-s) x + s$ one finds $$ I=\int_s^t (t-\sigma)^{N+1}(\sigma -s)^{N+1} \, d\sigma = (t-s)^{2N+3}\int_0^1 (1-x)^{N+1} x^{N+1}\, dx. $$ Now notice that $(1-x)x\le 1/4$ so that $$ I\le (t-s)^{2N+3} \frac{1}{4^{N+1}} = \frac{(t-s)^{2N+3}}{2^{2N+2}}. $$

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    Notice that $$\int_0^1 (1-x)^{\alpha-1} x^{\beta-1}\,dx = \mathrm{B}(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}.$$ – 2017-01-06
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    So, there actually is an equality if $t\ge s$. The request for an estimate was misleading. – 2017-01-06