Let $f: (0,\infty) \rightarrow [0,\infty]$ be Borel-measurable function. And let $a,b \in (0,\infty)$. We want to show that $$\int_{0}^{\infty} \int_{0}^{\infty} f(s+t)s^{a} \frac{ds}{s}t^{b} \frac{dt}{t} = B(a,b)\int_{0}^{\infty} f(u) u^{a+b} \frac{du}{u},$$ where $B(a,b) = \int_{0}^{1} v^{a-1} (1-v)^{b-1}dv$.
My attempt: I first try to start from the left-hand side. I first evalute $\int_{0}^{\infty} f(s+t)s^a \frac{ds}{s}$. I let $u = s+t$. So we have $du = ds$. We plug in and get \begin{align} \int_{0}^{\infty} f(s+t)s^{a}\frac{ds}{s} &= \int_{t}^{\infty} f(u)(u-t)^a\frac{du}{u-t}\\ &= \frac{(u-t)^a}{u-t} \int_{t}^{\infty} f(u)du. \end{align} So we have $$\int_{0}^{\infty}\int_{0}^{\infty} f(s+t)s^a\frac{ds}{s} t^{b} \frac{dt}{t} = \int_{0}^{\infty}\frac{(u-t)^a}{u-t} \int_{t}^{\infty} f(u)du t^b\frac{dt}{t}.$$ But I don't know how to go from here to relate it to $B(a,b)$. Please help! Thanks!
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1 Answer
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let us use the substitution $s=uv,t=u(1-v)$ we have that $u$ ranges from $0$ to $\infty$ while $v$ ranges from $0$ to $1$. Using this we get
\begin{align} \int_{0}^{\infty} \int_{0}^{\infty} f(s+t)s^{a-1} t^{b-1} \, ds \, dt=\int_{0}^{1} \int_{0}^{\infty} f(u)(uv)^{a-1} (u(1-v))^{b-1} \, udu \, dv=\\ \int_{0}^1 \int_{0}^{\infty} f(u) u^{a+b-1} v^{a-1} (1-v)^{b-1} \, du \, dv=B(a,b)\int_{0}^{\infty} f(u) u^{a+b-1} \, du.\end{align}
