Is Integration Greater than the Function We are Integrating?

Question

On p. 461 of the Sixth Edition of Hardy and Wright's An Introduction to the Theory of Numbers (Theorem 423). (see this post) it is written that - $$ \int_{n-1}^{n}\log^h{(\frac x t)}dt \geq \log^h{(\frac x n)} $$

My Understanding: We see that $\int_{n-1}^{n}f(x)\,dx$ is accumulating/summing/integrating all points from $n-1$ to $n$ for function $f$ whereas $\int_{n-1}^{n}f(n)\,dx = f(n)$ is just holding the value of $n$ for function $f$, thus it is always true that $$f(n)\leq \int_{n-1}^{n} f(t)dt $$

Is it correct?

Confusion: I have been told that it's generally true if $f$ is decreasing (and integrable), which $t\mapsto \log^h(x/t)$ is. If $f$ is decreasing, then $\int_{n-1}^n f(t)\textrm{d}t\geq \int_{n-1}^n f(n)\textrm{d}t=f(n).$ If $f$ is increasing, then $f(x) \le f(n)$ for all $x \in [n-1,n]$ Hence,$\int_{n-1}^{n}f(x)\,dx \le \int_{n-1}^{n}f(n)\,dx = f(n)$, but how does increasing or decreasing matter ? The function is given for a point and integration is done in between an interval.

Please clear my confusion in detail, in general,

How can I prove rigorously that -

$$f(n)\leq \int_{n-1}^{n} f(t)dt ?$$

what are the necessary conditions?

Answer 1

I'll just explain what they did there. Bro... Let $a = Re(s) > 0$. Then $\frac 1{t^{1+a}}$ is decreasing on $[n,n+1]$. So, $\frac 1{t^{1+a}}\le\frac 1{n^{1+a}}$ for $t\in [n,n+1]$. Thus, $\int_n^x\frac 1{t^{1+a}}\,dt\le\int_n^x\frac 1{n^{1+a}}\,dt$. Now, the integrand on the RHS is a constant, so you can pull it out of the interal: $$\int_n^x\frac 1{t^{1+a}}\,dt\le\int_n^x\frac 1{n^{1+a}}\,dt = \frac 1{n^{1+a}}\int_n^x\,1\,dt = \frac 1{n^{1+a}}(x-n)\le \frac 1{n^{1+a}},$$because $x-n\le 1$.

Answer 2

It is true in general that: $$ \int_a^b f(x)dx \ge \inf_{x\in(a,b)}(f(x))\cdot (b-a) $$ $\inf_{x\in(a,b)}(f(x))$ is basically the minimum value of $f(x)$ for $x\in(a,b)$, if you are unfamiliar with infimum.

For example, if $f$ is decreasing, then $f(x)\ge f(b)$ for all $x<b$. Then: $$ \int_a^b f(x)dx \ge \int_a^b f(b)dx = f(b)\cdot(b-a) $$

If f is increasing, then $f(x)\le f(b)$ for all $x<b$. Then: $$ \int_a^b f(x)dx \le \int_a^b f(b)dx = f(b)\cdot(b-a) $$

Think about the integral as the area under $f$. Changing $f(x)$ to the constant $f(b)$ means that we find the area of the rectangle with base $(b-a)$ and height $f(b)$. Depending on wether $f$ is increasing or decreasing, that rectangle will have greater or smaller area the the graph of $f$.