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I don't know how to put this properly but

Is there any some kind of equality or inequality like in case we know that $|\int_a^b f| \leq \int_a^b |f|$

$\Big[\int_a^b f(x) \Big]^{T}$ where $T \in \mathbb{R}^{+}$( positive real number) and $f(x)$ is positive for all $x$

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    $\begingroup$ Can you add some context or at least some informations about $f$ or $T$ ? $\endgroup$ – Tuvasbien Jan 19 at 20:33
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    $\begingroup$ I don't get the question either. Assuming that the integral is welldefined, it evaluates to a number, which you then take to a power. What kind of formula do you expect? $\endgroup$ – PrudiiArca Jan 19 at 20:39
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    $\begingroup$ I guess you are searching for a formula like $\left(\int_a^b f\right)^T=\Phi\left(\int_a^b f\right)$ but as far as I know, such a formula doesn't exist (or it involvs multiple integrals if $T$ is an integer, which is not very interesting). There exists inequalites such as Hölder's inequality. $\endgroup$ – Tuvasbien Jan 19 at 20:41
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    $\begingroup$ I see you edited your question. If you are searching inequalities, the most famous are Hölder's inequality : $\int f g\leqslant\left(\int f^p\right)^{1/p}\left(\int g^q\right)^{1/q}$ for all integers $p$ and $q$ such that $\frac{1}{p}+\frac{1}{q}=1$, Minkowski's inequality : $\left(\int (f+g)^p\right)^{1/p}\leqslant\left(\int f^p\right)^{1/p}+\left(\int g^p\right)^{1/p}$. Hölder's inequality gives you $\int_a^b f\leqslant \left(\int_a^b f^{T}\right)^{1/T}(b-a)^{1-1/T}$ and thus $\left(\int_a^b f\right)^T\leqslant (b-a)^{T-1} \left(\int_a^b f^{T}\right)$ $\endgroup$ – Tuvasbien Jan 19 at 20:49
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    $\begingroup$ It now certainly does (after edit) :) I apologize, in case I did appear grumpy in any way. $\endgroup$ – PrudiiArca Jan 19 at 20:52
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Jensen's inequality says that if $\varphi$ is convex then

$$ \varphi\left( \frac{1}{b-a} \int_a^b f \right) \le \frac1{b-a} \int_a^b \varphi \circ f.$$

The function $\varphi(x) = x^T$ is convex provided $T \ge 1$ so we have

$$ \frac{1}{(b-a)^T} \left( \int_a^b f \right)^T \le \frac1{b - a} \int_a^b f^T, \quad 1 \le T. $$

Which when we rearrange is

$$ \left( \int_a^b f \right)^T \le (b - a)^{T - 1} \int_a^b f^T, \quad 1 \le T. $$

If $b - a \le 1$ then this implies the weaker inequality

$$ \left( \int_a^b f \right)^T \le \int_a^b f^T, \quad 1 \le T. $$

Also, if $0 \le T \le 1$ the function is concave so we have the opposite inequality:

$$ \frac{1}{(b - a)^T} \left( \int_a^b f \right)^T \ge \frac1{b- a} \int_a^b f^T, \quad 0 < T \le 1. $$

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  • $\begingroup$ @user64742 The question is what sort of inequality exists between $\int f^T$ and $(\int f)^T$. That is what I have provided and that is what the person asking the question has accepted. If this did not answer their question they would not have accepted it. $\endgroup$ – Trevor Gunn Jan 23 at 0:27
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In general the statement is not true. A counter-example is found by taking $a = 0, b= 15, f(x) = x^2$ and $T = 2$, since then

\begin{align*} \mathopen{}\left[\int_{a}^{b}f(x)\,\mathrm{d}x \right]^{T}\mathclose{} &= \mathopen{}\left[\int_{0}^{15}x^{2}\,\mathrm{d}x\right]^{2}\mathclose{} \\[1ex] &=1\,265\,625 \end{align*} while \begin{align*} \int_{a}^{b}\Big[ f(x)\Big]^{T}\,\mathrm{d}x &= \int_{0}^{15}x^{4}\,\mathrm{d}x \\[1ex] &=151\,875. \end{align*}

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