Let $f(x)$ be continuous on $[0,1]$ with $\int_0^1 f(x)dx=\alpha$. Find $\int_0^1 \int_x^1 f(x)f(y)dydx$.

I don't really know what I'm doing at all with this one. I started off by letting $F(x)$ be an antiderivative of $f$, and working out the inner integral as $\int_x^1 f(x)f(y)dy=f(x)(F(1)-F(x))$. But that's the only thing I can come up with, and I think it doesn't really get anywhere, because I then have to integrate that expression with respect to x, which doesn't seem possible.

Any pointers?



Note that the integrand is symmetric in $x,y$; so can you find a relation between this integral and the integral $\int\limits_0^1\int\limits_0^1 f(x)f(y)\,\mathrm dx\,\mathrm dy$, the latter of which easily evaluates to $\alpha^2$

Think of the analogous case in summation: what is the relation between $\sum_{i=0}^n\sum_{j=i}^n f(i,j)$ and $\sum_{i=0}^n\sum_{j=0}^n f(i,j)$ when $f(i,j)=f(j,i)$ ?


Starting from where you left off, we need to integrate:

$$I=\int_{0}^{1}f(x)(F(1)-F(x))\;dx$$ $$=\int_{0}^{1}f(x)F(1)\; dx-\int_{0}^{1} f(x)F(x)\;dx$$ $$=F(1)\underbrace{\int_{0}^{1}f(x)\; dx}_{\alpha}-\underbrace{\int_{0}^{1} f(x)F(x)\;dx}_{J}$$

For $J$, recall that $F'(x)=f(x)$ so that:


and let:

$$u=F(x),\quad du=F'(x)\; dx$$

so that:



(where we used $a^2-b^2=(a-b)(a+b)$)

Note that $\alpha=F(1)-F(0)$ and so substituting this result we get:


and substituting this in what we found for $I$:

$$I=\alpha F(1)-\frac{1}{2}\alpha(F(1)+F(0))$$ $$I=\frac{\alpha}{2}( F(1)-F(0))$$ And again using $\alpha=F(1)-F(0)$:


(Edits: spelling and layout)


\begin{align*} \int_{0}^{1}\int_{x}^{1}f(y)f(x)dydx&=\int_{0}^{1}\left(\alpha-\int_{0}^{x}f(y)dy\right)f(x)dx\\ &=\alpha\int_{0}^{1}f(x)dx-\int_{0}^{1}\int_{0}^{x}f(y)f(x)dydx\\ &=\alpha^{2}-\int_{0}^{1}\int_{y}^{1}f(y)f(x)dxdy\\ &=\alpha^{2}-\int_{0}^{1}\int_{x}^{1}f(x)f(y)dydx, \end{align*} so \begin{align*} \int_{0}^{1}\int_{x}^{1}f(y)f(x)dydx=\dfrac{\alpha^{2}}{2}. \end{align*}


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