Using Fubini to prove $p\int_t^\infty\frac1{s^{p+1}}(\int_0^s w(x) \, dx) \, ds=\int_t^\infty\frac{w(s)}{s^p} \, ds+\frac{\int_0^tw(x) \, dx}{t^p}$

Question

It was written that using Fubini's Theorem for a positive function $w$ and $p>0$, the following relation holds \begin{equation*} p\int_t^\infty \frac{1}{s^{p+1}}\left( \int_0^s w(x) \, dx\right) \, ds=\int_t^\infty \frac{w(s)}{s^p} \, ds+\frac{\int_0^t w(x) \, dx}{t^p}.\tag1 \end{equation*}

My try for this, as follows by applying Fubini's Theorem

\begin{eqnarray*} p\int_t^\infty \frac{1}{s^{p+1}}\left( \int_0^s w(x) \, dx\right)\, ds &=&p\int_t^\infty w(x)\left( \int_t^\infty \frac{1}{s^{p+1}} \, ds\right) dx \\ &=&\int_t^\infty w(x)\left. \frac{1}{s^p}\right\vert_\infty^t \, dx \\ &=&\int_t^\infty w(x)\frac{1}{t^p} \, dx. \end{eqnarray*}

Which obviously gives only one term from the required relation $(1).$ Any help with this, please.

Answer 1

\begin{align} & \int_t^\infty \frac{1}{s^{p+1}}\left( \int_0^s w(x) \, dx\right) \, ds \\[12pt] = {} & \iint\limits_{x,s\,:\,t\,<\, s\ \&\ 0\,<\,x\,<\, s} \frac 1 {s^{p+1}} w(x) \, d(x,s) \\[12pt] = {} & \iint\limits_{\big(x,s\,:\,t\,<\,x\,<\,s\big)} + \iint\limits_{\big( x,s\,:\,0\,< \,x\, < \,t\,<\,s\big)} \end{align}

Answer 2

First of all, you wrote LHS in two different ways (messing with $x$ and $t$).

The correct solution should look something like this $$p\int\limits_t^\infty\frac{1}{s^{p+1}}\left( \int\limits_0^s w(x)dx \right) ds=p\int\limits_0^t w(x) \left( \int\limits_t^\infty \frac{1}{s^{p+1}} ds \right)dx+ p\int\limits_t^\infty w(x) \left( \int\limits_x^\infty \frac{1}{s^{p+1}} ds \right)dx.$$

Simplifying these two integrals (the same way you did in the question body) will give you the right identity.

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(edit)

You asked for details in a comment. Fubini's theorem allows us to compute a double integral over some domain as an iterated integral. In your problem we are given an iterated integral. It gives us a domain over which a double integral is taken. In the picture it is the red domain $D$. It looks like that, because we know that $s$ must run from $t$ to $+\infty$ and $x$ must run from $0$ to the line given by $s=x$. Since double integral is additive, we can write $\iint\limits_D\dots=\iint\limits_A\dots+\iint\limits_B\dots$, where $A$ and $B$ are respectively blue and green domains from the picture. Now for each of them we use Fubini's theorem: domain A is obtained when $x$ runs from $0$ to $t$ and $s$ runs from $t$ to $+\infty$. It gives us the first iterated integral in the RHS. Domain $B$ is obtained when $x$ runs from $t$ to $+\infty$ and $s$ runs from line $s=x$ to $+\infty$. It gives us the second iterated integral in the sum in the RHS. Is the solution clear now?