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I'm quite stuck with the following practice question. Suppose $f,g: R \to R$ are continuous $2\pi$-periodic functions. Let $h(s) =\int_{0}^{2\pi} f(s-t)g(t) \ dt$. Prove that

$$\int_{0}^{2\pi} h(s) \ ds = \left( \int_{0}^{2\pi} f(t)\ dt \right) \left( \int_{0}^{2\pi} g(t) dt \right)$$

My Attempt

$\begin{equation} \int_{0}^{2\pi} h(s) \ ds = \int_{0}^{2\pi}\int_{0}^{2\pi} f(s-t)g(t) \ dt \ ds \\ = \int_{0}^{2\pi}g(t) [\int_{0}^{2\pi} f(s-t)\ ds] \ dt \end{equation}$

Could someone point out the right direction to go from here?

2 Answers2

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Consider changing the bounds of integration on the inner integral, in order to shift the variable under consideration (your goal is to have something like $f(r)$ on the inside of the integral). When the variable shifts, the bounds must shift the other way in order to reflect that change.

For example, $\int_{x=1}^3 x+y\ dx = \int_{r=y+1}^{y+3}r\ dr$.

When dealing with the inner integral, remember that you can think of $t$ as if it's a constant (just imagine $t=1$ or something like that). After that, it should be possible to use the periodic nature of the functions.

Adina Goldberg
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    I've tried doing $\int_{-t}^{2 \pi - t} f(s) \ ds$, but that hasn't lead anywhere. Above all I'm confused how I'm meant to change $ds$ to $dt$ –  Jul 28 '20 at 19:16
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    @userhello108 The integral of any periodic function over their period can be shifted to a different interval of period length at no penalty. Since you have the $t$ fully in the bounds now you are free to remove them i.e. $$\int_{-t}^{2\pi-t} f(s):ds = \int_0^{2\pi}f(s):ds$$ – Ninad Munshi Jul 28 '20 at 19:19
  • @userhello108 It might help to name your new variable something new! So let's say we set $r = s - t$. Then your bounds are correct, and what's left is to think about what $ds$ should become in terms of $dr$. If you've seen change of variables before, you'd have seen that $ds$ can be replaced by $\frac{ds}{dr} dr$. So you'd need to compute a (simple) derivative to finish up, by expressing $s$ as a function of $r$ with some rearranging and then taking the derivative. – Adina Goldberg Jul 28 '20 at 19:20
  • Another thing that may be helpful is that in the expression $(\int f(t)\ dt) (\int f(s)\ ds)$, changing the $s$ to a $t$ does not change the value of the expression, as each integral is computed separately and then they are multiplied together. – Adina Goldberg Jul 28 '20 at 19:22
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\begin{align} &\Rightarrow\int_{0}^{2\pi} h(s) \ ds \\ &= \int_{0}^{2\pi}\int_{0}^{2\pi} f(s-t)g(t) \ dt \ ds\\ &= \int_{0}^{2\pi}\int_{0}^{2\pi} f(s-t)g(t)\ ds \ dt&(\text{Changing the order of integration})\\ &= \int_{0}^{2\pi}g(t) \left(\int_{0}^{2\pi} f(s-t)\ ds \right)\ dt&(\text{Separating the independent integrand})\\ &= \int_{0}^{2\pi}g(t) \left(\int_{0-t}^{2\pi-t} f(s)\ ds\right) \ dt&(\text{Shifting property})\\ &= \int_{0}^{2\pi}g(t) \left(\int_{0}^{2\pi} f(s)\ ds\right) \ dt&(\text{Periodicity})\\ &= \left(\int_{0}^{2\pi} f(s)\ ds\right) \left(\int_{0}^{2\pi}g(t) \ dt\right)&(\text{Separating independent integrals})\\ &= \left(\int_{0}^{2\pi} f(t)\ dt\right) \left(\int_{0}^{2\pi}g(t) \ dt\right)&(\text{Changing the variable name}) \end{align}

Sameer Baheti
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