Proving the inequality

$$\int_{1}^{b}a^{\log _bx}dx > {\ln b} : a,b>0 , b\neq1$$

Solving the integral, I found the result below,

$$\frac{\ ab-1}{{\ln ab}}{\ln b}$$

I know that I need only prove that, this part $$\frac{\ ab-1}{{\ln ab}}$$ must be greater than $1$, which I am unable to proceed further.

I ran to a difficulty when, $a= \frac{1}{b}$ or $b= \frac{1}{a}$ as $b$ ${\neq1}$ Please help.

Any alternative way to prove the inequality will be valued.

  • $\begingroup$ The inequality you try to prove is not true. Could you please verify that you wrote the problem correctly? $\endgroup$
    – mickep
    Mar 12, 2017 at 10:41
  • $\begingroup$ I believe the inequality is reversed for values of b such that $0<b<1$ $\endgroup$ Mar 12, 2017 at 12:44

3 Answers 3


Your inequality isn't true for all $a,b>0$ with $b\neq 1$. Here's a proof for the domain where it is true: $$\{(a,b)\in\mathbb{R}^+:ab>1\land b>1\}\cup\{(a,b)\in\mathbb{R}^+:ab<1\land b<1\}$$

The key is that this is the region in $\frac{\ln(b)}{\ln(ab)}>0$ So we can multiply and divide inequalities by that without changing the sign. You have that you want to prove that


Since $\frac{\ln(b)}{\ln(ab)}>0$ we can divide both sides of the I equality by that. This gives $ab>1+\ln(ab)$ or, making the substitution $x=ab$, $x>1+\ln(x)$ and then exponentiating gives $e^x>ex\Rightarrow e^{x-1}>x$. Take the power series for $e^z$ and plug in $x-1$, to get


Simplifying gives $$\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots>0$$

Which holds for all $x$ because it's either a strictly positive series or an alternating series that doesn't converge to zero with positive first term. All of these implications are if and only if statements, so the original inequality holds.

Outside of the specified region, the ratio $\frac{\ln(b)}{\ln(ab)}$ is negative, and so following this exact same line of reasoning proves that the reverse inequality is true outside of the specified region. The only case not addressed here is when $b=1$, in which case the inequality is false and instead equality holds.

  • $\begingroup$ I don't think $x=ab$ because b can't be equal to 1. Also if $a= \frac{1}{b}$ what do we do then? $\endgroup$ Mar 12, 2017 at 7:09
  • $\begingroup$ $x>1+\ln(x)$ is invalid for $x=1$. I think the original problem is incorrect, and needs a $\ge$ instead of a $>$. $\endgroup$ Mar 12, 2017 at 7:55
  • $\begingroup$ @jyotishrajthoudam Why would $b\neq 1$ mean we can't do the substitution $x=ab$? If $ab=1$ then the expression for the integral that you d calculated is $0/0$ so at some point during the integration you probably assumed that $ab\neq 0$ $\endgroup$ Mar 12, 2017 at 12:15
  • $\begingroup$ @AkivaWeinberger I don't think so. That solves the problem for $x>1+\ln(x)$ but in the expression $(x-1)/\ln(x)$ plugging in $x=1$ gives $0/0$ $\endgroup$ Mar 12, 2017 at 12:16
  • $\begingroup$ @StellaBiderman The limit turns out to be 1, perhaps that's what they meant? $\endgroup$ Mar 12, 2017 at 12:18

I leave this as an answer, since it is the only way for me to include an image. I think the problem in the question is invalid, and should be updated to be correct (or removed if the true problem in itself is as stated).

I let Mathematica plot the region where $$ \ln b\frac{ab-1}{\ln(ab)}<\ln b. $$ It is the blue domain below.

domain of failure

For example, for $a=1/2$ and $b=4/3$ one gets $$ \ln b\frac{ab-1}{\ln(ab)}-\ln b\approx -0.05. $$

  • $\begingroup$ Thanks for the image! This helped me realize what the issue was and exactly when the inequality holds. $\endgroup$ Mar 12, 2017 at 16:40

As every other part of the answer has been cross-examined, I'm going to show the integration and hopefully you guys can spot what's gone wrong.

$$\mathrm I = \int_{1}^{b}a^{\log _bx}dx $$

After the substitution $x=b^u \implies dx=b^u \mathrm{ln}(b)du$ which is a nice continuous and differentiable function, we reduce the integral to $$ \int_0^1 a^u \times b^u \mathrm{ln}(b)du $$ as $x=1 \implies u=0 $ and $ x=b \implies u=1 $

Simplifying to $$ \mathrm I = \mathrm{ln}(b) \int_{0}^{1} (ab)^udu $$

As $\frac{(ab^u)}{\mathrm{ln}(ab)}$ is a primitive of $(ab)^u$ (Assuming $ab \ne 1$)and is continuous and differentiable we can apply the Newton-Leibnitz formula to give us $$ \mathrm I = \frac{\ ab-1}{{\ln ab}}{\ln b} $$

So Its understandable why the case $ab=1$ is problematic. Note that evaluating the integral for the case $ab=1$ returns an equality, contradicting the conclusion of the problem

The other possibilities may stem from the fact that while $ ab-1 > \mathrm{ln}(ab) $ The statement $ \frac{\ ab-1}{{\ln ab}} >1 $ also requires $ab>1$

Similarly the statement $ \frac{\ ab-1}{{\ln ab}} >1 \implies \ln(b) \frac{\ ab-1}{{\ln ab}} >ln(b)$ iff $b>1$

I see no reason to think that any assumptions have been made during the integration to exclude the previous two cases, If you do please comment.

  • $\begingroup$ The $I$ will not be the same as before when $b<1$ the limits has to flip and also the $ln b$ won't be positive. How to deal with that case? $\endgroup$ Mar 30, 2017 at 17:49

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