Largest natural number $n$ such that $\int_1^\infty\frac{\log^nx}{e^x}\mathrm dx<1$

Question

Find the largest natural number $n$ such that $$\int_1^\infty\frac{\ln^nx}{e^x}\mathrm dx<1$$

Upon searching on site and google, I found a similar integral:

$$\int_0^\infty\frac{\ln x}{e^x}\mathrm dx=-\gamma$$

But the techniques involving solving such integrals is beyond what I've learnt till now. The elementary method I know is to think of some adaptation of squeeze theorem like to search a $f$ such that: $$f>\frac{\ln^nx}{e^x}$$ and $$\int_1^\infty f\mathrm dx = 1$$ But I can't imagine any such $f$ upon hit and trial.

(P.S.: This question came in my class test of applications of derivatives. According to answer key, $n=7$)

Answer 1

Some helps/remarks (hope you enjoy it):

Using Bernoulli's inequality we have :

$$J=\int_{1}^{\infty}\frac{\left(1+4\left(\ln\left(x\right)-1\right)\right)^{2}}{e^{x}}dx< I_8$$

Using a simple bound we have :

$$\int_{1}^{\infty}\frac{\left(1+4\left(\frac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-1\right)\right)^{2}}{e^{x}}dx<J$$

or :

$$\int_{1}^{\infty}\frac{\left(-3+\frac{16\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\right)^{2}}{e^{x}}dx<J$$

or :

$$K=\int_{1}^{\infty}\frac{9+\frac{16^{2}\left(\sqrt{x}-1\right)^{2}}{\left(\sqrt{x}+1\right)^{2}}-6\cdot16\frac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}}{e^{x}}dx$$

Now we have :

$$\int_{1}^{\infty}\frac{9+\frac{29\left(x-1\right)}{\left(x+1\right)}-6\cdot16\frac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}}{e^{x}}dx<K$$

finally :

$$1<\int_{1}^{\infty}\frac{9+\frac{29\left(x-1\right)}{\left(x+1\right)}-51\frac{\left(x-1\right)}{\left(x+1\right)}}{e^{x}}dx<K$$

It shows that it cannot be greater than $8=n$

Answer 2

I do not know a simple solution to this problem.

Consider $$I_n=\int_1^\infty e^{-x} \,\log ^n(x)\,dx$$ For $n>1$, the explicit solution is gives as a Meijer_G function (which can express in terms of generalized hypergeometric functions).

$$\color{blue}{I_n=n!\, G_{n,n+1}^{n+1,0}\left(1\left| \begin{array}{c} 1,1,1,\cdots,1 \\ 0,0,0,0,\cdots,0 \\ \end{array} \right.\right)}$$

For sure, this leads to the result.