# power of integral

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 $$x$$

• Can you add some context or at least some informations about $f$ or $T$ ? – Tuvasbien Jan 19 at 20:33
• 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? – PrudiiArca Jan 19 at 20:39
• 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. – Tuvasbien Jan 19 at 20:41
• 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 $\left(\int_a^b f\right)^T\leqslant (b-a)^{T-1} \left(\int_a^b f^{T}\right)$ – Tuvasbien Jan 19 at 20:49
• It now certainly does (after edit) :) I apologize, in case I did appear grumpy in any way. – PrudiiArca Jan 19 at 20:52

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:

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

• @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. – Trevor Gunn Jan 23 at 0:27

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*}