# Prove that the integral $\int_0^1 f(t) P(t - x)dt$ is a polynomial in $x$

So suppose $f$ is a generally complex, continuous function on $[0,1]$ and $P$ is a polynomial defined on the real numbers. Evidently $$\int_0^1 f(t)P(t - x)\,dt$$ is a polynomial in $x$ but that this is the case is not obvious to me. How might one go about proving this? Note that $f$ is continuous and so is integrable but it may not have a nice anti-derivative.

• The intuitive answer I would give a student would be that the $t$ terms disappear. Remember that definite integrals with respect to a given variable $\theta$ turn all instances of $\theta$ into constants, so all that is left is a polynomial in $x$ (and integration of a polynomial yields another polynomial). Of course, this explanation doesn't quite explain that the $t-x$ within the polynomial function leaves the function a polynomial once integrated, though this too can be reasoned by a similar thought process and the fact that integrating a polynomial yields another polynomial Jun 29 '16 at 2:33
• Additionally, the proof by Alozizio below should help in piecing together the above logic and provides a more rigorous foundation for the claim that $x \to t-x$ doesn't change the fact we get a polynomial in $x$ Jun 29 '16 at 2:36
• @BrevanEllefsen: The intuitive answer you gave is rigorous and is almost devoid of mathematical symbols. I like such arguments because they are easier to understand and do not compromise on rigor. Jun 29 '16 at 6:49

$P(X)=a_0+a_1X+\cdots+a_nX^n.$
$P(t-X)=a_0+a_1(t-X)+\cdots+a_n(t-X)^n=a_0+a_1t-a_1X+a_2t^2+2a_2tX+a_2X^2+\cdots$
$\implies f(t)P(t-X)=f(t)a_0+tf(t)-a_1X+t^2f(t)a_2+2a_2tf(t)X+f(t)a_2X^2+\cdots$
The above is meant for you to visualize what is happening. When you integrate with respect to $t$, everything alongside the $X$'s will turn to constants.
Alternatively, show that $Q(x):=\int_0^1\,f(t)\,P(t-x)\,\text{d}t$ satisfies $\frac{\text{d}^k}{\text{d}x^k}\,Q(x)=0$ for some integer $k>0$. As $P$ is a polynomial, $\frac{\partial^k}{\partial x^k}\,P(t-x)=0$ if $k:=1+\deg(P)$. Thus, the Leibniz Integral Rule gives $$\frac{\text{d}^k}{\text{d}x^k}\,Q(x)=\int_0^1\,f(t)\,\frac{\partial^k}{\partial x^k}\,P(t-x)\,\text{d}t=0\,.$$ This result holds for any integrable function $f:[0,1]\to\mathbb{C}$.