For $0<\theta, \lambda<1$ and $c>1$, we wish to upper bound the following gamma function: $$\int_{\theta}^{1} t\exp \left(-c\left(\lambda t+\frac{1}{t}\right) \right)dt$$


You don't say that you need an optimal bound, so I just give a bound. I assume that $0<\lambda<1$ and $0<\theta<1$.

Using the arithmetic-geometric inequality, $$ \lambda t+1/t\geq 2\sqrt{\lambda}. $$ Hence $$ \exp(-c(\lambda t+1/t))\leq \exp(-2c\sqrt{\lambda}), $$ and the integral is bounded by $$ \exp(-2c\sqrt{\lambda})(1-\theta^2)/2. $$

Slightly better:

Instead of using AM-GM inequality, since $\lambda<1$ the minimum of $\lambda t+1/t$ is attained at $t=1$, so $$ \lambda t+1/t\geq \lambda+1. $$ The rest is as before, and your integral is bounded by $$ \exp(-c(\lambda+1))(1-\theta^2)/2. $$

  • $\begingroup$ thanks for your response! Could it be further tighten. $\endgroup$ – Ram Sep 4 '15 at 3:57
  • $\begingroup$ In fact, it seems that it is pretty tight as it is. It could of course be better, in particular for $c\approx 1$. But for $c>5$, say, it is pretty good. I think you have to tell more explicitly what you need. $\endgroup$ – mickep Sep 4 '15 at 10:24

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