A well known result of definite integrals is:
$$\int_{0}^{a} f(x)dx=\int_{0}^{a} f(a-x)dx$$
It is very easy to verify this result by substituting $x=a-y \implies dx=-dy$.
When $x=0, y=a$, and when $x=a, y=0$. Substituing this in $\int_{0}^{a}f(x)dx$, we get $\int_{a}^{0}f(a-y)(-dy)$. The minus sign in the integrand can be used to reverse the limits. Also we can use $x$ instead of $y$.
My question is:
If we know that the value of $\int_{0}^{a}f(x)dx=I$, is there a formula to evaluate $$\int_{0}^{a}f(x)f(a-x)dx$$ in terms of $I$?
I tried to use the above result, but could not conclude anything.
Take this as an example;
We know that $\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4}$, is it possible to evaluate (using a formula in terms of $(\pi/4)$)
$$\int_{0}^{1}\frac{1}{1+x^2}\cdot\frac{1}{1+(1-x)^2}dx$$
in terms of $\frac{\pi}{4}$, without using partial fractions decomposition?
It may be helpful to consider only the case when $f:[-R,R]\rightarrow\mathbb{R}$ is real-analytic on $(-R,R),$ and we let $a\in(0,R),$ for simplicity. In that case, there exists some $(a_n)_{n\in\mathbb{N}}$ such that $$f(x)=\sum_{n=0}^{\infty}a_nx^n,\;x\in(-R.R).$$ Under that consideration, $$f(a-x)=\sum_{n=0}^{\infty}a_n(a-x)^n$$ and $$\int_0^af(x)f(a-x)\,\mathrm{d}x=\sum_{n=0}^{\infty}a_n\int_0^a(a-x)^nf(x)\,\mathrm{d}x.$$ For $n=0,$ the integral $$\int_0^a(a-x)^nf(x)\,\mathrm{d}x$$ is trivial, and is equal to $I.$ However, it is more complicated for $n\gt0.$ By Leibniz' integral rule, we have that $$\frac{\partial}{\partial{a}}\int_0^a(a-x)^nf(x)\,\mathrm{d}x=(a-a)^nf(a)+\int_0^a\frac{\partial}{\partial{a}}(a-x)^nf(x)\,\mathrm{d}x=n\int_0^a(a-x)^{n-1}f(x)\,\mathrm{d}x.$$ As such, one can define $$I_n(a)=\int_0^a(a-x)^nf(x)\,\mathrm{d}x,$$ where $$I_0(a)=I,$$ thus having that $$I_n'(a)=nI_{n-1}(a).$$ Also, notice that $$I_n(0)=0$$ for all $n\in\mathbb{N}.$ Hence $$I_n(a)=n\int_0^aI_{n-1}(s)\,\mathrm{d}s$$ or $$I_{n+1}(a)=(n+1)\int_0^aI_n(s)\,\mathrm{d}s,$$ which is the perfect recursion to solve for $I_n(a)$ in terms of $I_0(a)=I.$ This implies that $$I_{n+2}(a)=(n+2)(n+1)\int_0^a\int_0^tI_n(s)\,\mathrm{d}s\,\mathrm{d}t.$$ If we adopt the convention of defining a linear operator $J$ defined by $J[g]=\int_0^ag(x)\,\mathrm{d}x,$ then we in general have that $$I_{n+m}(a)=\frac{(n+m)!}{n!}(J^m)[I_n](a),$$ which can be proven by simply applying the principle of induction. Letting $n=0,$ it would result in $$I_m(a)=m!(J^m)[I_0](a).$$ Thankfully, we have a formula for repeated integration by Cauchy: $$(J^m)[g]=\frac1{(m-1)!}\int_0^a(a-x)^{m-1}g(x)\,\mathrm{d}x.$$ Therefore, $$I_n(a)=n\int_0^a(a-x)^{n-1}I_0(x)\,\mathrm{d}x.$$ Therefore, $$\int_0^af(x)f(a-x)\,\mathrm{d}x=\sum_{n=0}^{\infty}a_nI_n(a)=\sum_{n=0}^{\infty}na_n\int_0^a(a-x)^{n-1}I_0(x)\,\mathrm{d}x.$$ This is by no means a closed-form formula, though, as this involves an abstract integral within an infinite series.