# Inequality for the integral $\frac{\ln x}{x^n}$

Define the integral $I_{n}$ as follows for $n$ an integer greater than $1$: $I_{n}:=\int_{1}^{e}\frac{\ln x}{x^n}dx$

Is it true that

$$I_{n}\leq \frac{1}{n-1}\left(1-\frac{1}{e^{n-1}}\right)?$$

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Yes.    – Did Dec 31 '11 at 14:28
This integral is easy to evaluate using integration by parts (set $u=\ln x$). – David Mitra Dec 31 '11 at 14:40

For $n>1$:
$$\int_1^e{1\over x^n}\,dx= {x^{-n+1}\over -n+1}\biggl|_1^e={e^{-n+1}\over -n+1} -{1\over -n+1}={1\over n-1}(1-e^{-n+1}).$$
Now use $0\le\ln x\le 1$ for $1\le x\le e$ to obtain $$\int_1^e{\ln x\over x^n}\,dx\le \int_1^e{1\over x^n}\,dx={1\over n-1}(1-e^{-n+1}).$$
Or, you can evaluate the integral exactly: \eqalign{ \int_1^e \underbrace{(\ln x)\vphantom{x^n\over x^n1}}_u \,\,\underbrace{x^{-n}\,dx\vphantom{1\over x}}_{dv} &= \underbrace{\ln x\vphantom{x^n\over1 x^n}}_u\,\,\underbrace{ {x^{-n+1}\over -n+1}}_{v}\biggl|_1^e - \int_1^e \underbrace{{x^{-n+1}\over -n+1}}_v\,\,\underbrace{ {1\over x}\,dx}_{du}\cr &={ e^{-n+1}\over -n+1} +{1\over n-1}\int_1^e x^{-n}\,dx\cr &={ e^{-n+1}\over -n+1} +{1\over n-1}\cdot {1\over n-1}(1-e^{-n+1})\cr &={1\over n-1}\Bigl( -e^{-n+1} +{1\over n-1}(1-e^{-n+1})\Bigr)\cr &={1\over n-1}\Bigl( {1\over n-1} -{n\over n-1} e^{-n+1}\Bigr)\cr (&\le{1\over n-1}\Bigl( 1 - e^{-n+1}\Bigr).)\cr }