$\int_{e}^{\infty} \frac{x+1}{x^3\ln ^2(x)} dx$

Question

I am trying to solve the given integral (on my own, not a homework problem.)

$$\int_{e}^{\infty} \frac{x+1}{x^3\ln ^2(x)} dx$$

I am having problem with a part of using Feynman's technique which I need resolved.

I can rewrite this as:

$$\int_{e}^{\infty} \frac{x^{-2}+x^{-3}}{\ln ^2(x)} dx$$

And can create the function:

$$I(s,t) = \int_{e}^{\infty} \frac{x^{-s}+x^{-t}}{\ln ^2(x)} dx$$

Note that our original integral is $I(2,3)$. Here's where my confusion first sets in. I am attempting to get rid of the $\ln^2(x)$, so I can either compute $I_{tt} (s,t), I_{ss}(s,t),$ or $I_{st} (s,t)$, and these will ALL get rid of the $\ln^2(x)$ in the denominator. The first confusion is why this is. Here's my working after computing $\frac{\partial ^2I}{\partial s \partial t}$.

$$\frac{\partial ^2I}{\partial s \partial t}= \int_{e}^{\infty} (x^{-s} + x^{-t})dx = \frac{e^{1-t}}{1-t} + \frac{e^{1-s}}{1-s}$$.

$\int \frac{e^x}{x} dx$ doesn't have an elementary antiderivative, so call this integral $\text{Ei} (x)$, as it is often named.

Integrate with respect to $t$ yields:

$$\frac{\partial I}{\partial s} = -\text{Ei} (1-t) + \frac{te^{1-s}}{1-s} + f_1 (s)$$

Integrate with respect to $s$.

$$I(s,t) = -s\text{Ei} (1-t) - t\text{Ei} (1-s) + \int f_1(s) ds + f_2(t).$$

Here is my other question, which relates to Feynman's technique with a multivariable function in general, not necessarily with this question. When we differentiate under the integral sign to evaluate an integral, we use one parameter, and use a clever value for the parameter to evaluate the constant $C$ after integrating. When we integrate and there's partials involved, we have to introduce a whole FUNCTION with the other variable. How can we find this, and how in general can we use $I(s,t)$ to evaluate the integral? You can't simply put a nice value of $s$ or $t$ for a function.

Answer 1

Concerning the first integral, let $x=e^t$ to end with $$I=\int \frac{x+1}{x^3\log ^2(x)} dx=\int \frac{e^{-2 t}}{t^2} \,dt+\int \frac{e^{-t}}{t^2} \,dt$$ Using one integration by parts $$\int \frac{e^{-k t}}{t^2} \,dt=-k\, \text{Ei}(-k t)-\frac{e^{-k t}}{t}$$ This makes $$I=-\frac{2 t \,\text{Ei}(-2 t)+t \,\text{Ei}(-t)+e^{-2 t}+e^{-t}}{t}$$ Integrating from $1$ to $\infty$, this gives $$\int_{e}^{\infty} \frac{x+1}{x^3\log ^2(x)} dx=2\,\text{Ei}(-2)+\text{Ei}(-1)+\frac{1}{e^2}+\frac{1}{e}$$

You could also make the problem more general using, if $k>0$, $$\int t^{-n} \,e^{-kt}\,dt=-k^{n-1} \Gamma (1-n,k t)$$ $$\int_1^\infty t^{-n} \,e^{-kt}\,dt=k^{n-1} \Gamma (1-n,0)$$