1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $$ \exp(-c(\lambda+1))(1-\theta^2)/2. $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.