At $\lambda \to \infty $ Method of steepest descent approximate the integral
$$I(\lambda )=\int\limits_{1}^{\infty }\left ( \frac{\ln x}{x} \right )^{\lambda }dx$$
My attempt:
The method itself and the general formula
$$I(\lambda) = \int\limits_a^b f(x) e^{\lambda S(x)} \, dx, \quad \lambda \to \infty.$$
где $f(x)$ и $S(x)$ – некоторые функции.
I have not been able to apply this method to my integral. At $x>e$ the function decreases, and decreases more rapidly as $\lambda$ increases. Consequently, most of the contribution to the integral at $\lambda \to \infty$ comes from values of $x$ close to $e$. In the limit at $\lambda \to \infty$ the integral will tend to zero, i.e.
$$I(\lambda )=\int\limits_{1}^{\infty }\left ( \frac{\ln x}{x} \right )^{\lambda }dx \to 0, \quad \lambda \to \infty$$
This is my reasoning, but I have not been able to calculate an approximate integral. Could you help me?
Best Answer
The integral can be studied with Laplace's method. One has
$$ I = \int_1^{\infty}\left(\frac{\ln x}{x} \right)^{\lambda} dx = \int_1^{\infty} \exp\left(-\lambda \left(\ln x -\ln \ln x\right)\right)dx, \quad \lambda \rightarrow \infty $$
Let $p(x) = \ln x -\ln \ln x$, a minimum occurs around $x=e$ and Taylor expanding around this point, one has
$$ p(x) = 1 + \frac{1}{2e^2} (x-e)^2 - \frac{5}{6e^3}(x-e)^3 +... $$
Split the integral into two parts, the first one from $1$ to $e$ ($I_1$) and the second from $e$ to infinity ($I_2$). For the second integral, one may identify from the link the parameters $p(a) = 1,P=(2e^2)^{-1},\mu=2,Q=1,\lambda=1$, at the point $a=e$, which gives the dominating term
$$ I_2 = \int_1^{\infty}\left(\frac{\ln x}{x} \right)^{\lambda} dx \sim \frac{Q}{\mu}\Gamma\left(\frac{\lambda}{\mu} \right) \frac{e^{-\lambda p(a)}}{(Px)^{\lambda/\mu}} = e\sqrt{\frac{{\pi}}{2}} \frac{e^{-\lambda}}{\lambda^{1/2}} $$
One may similarly show that $I_1 = I_2$ which then gives
$$ I = \int_1^{\infty}\left(\frac{\ln x}{x} \right)^{\lambda} dx \sim \sqrt{2\pi}e\lambda^{-1/2}e^{-\lambda}, \quad \lambda \rightarrow \infty $$
to leading order. From the same link, it is shown how to include additional terms if necessary.