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Let $f$ be continous on [-a,a]

a) prove : $\int^{a}_{-a} f(x) dx = 0$

Because $f$ is odd $f(-x) = -f(x)$

$$\int_{x=-a}^0 f(x) \, dx = \int_{x=-a}^0 -f(-x) \, dx$$

letting $t = -x, dt = -dx$

$\int_{x=-a}^0 -f(-x) \, dx = \int_{t=a}^0 f(t) \, dt$

$\int_{t=a}^0 f(t) \, dt = -\int_{t=0}^a f(t) \, dt$

$\int_{-a}^{a}f(x)dx=\int_{-a}^{0}f(x)dx+\int_{0}^{a}f(x)dx$

$\int_{-a}^{a}f(x)dx=-\int_{0}^{a}f(x)dx+\int_{0}^{a}f(x)dx$=0

b) prove : $\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx$

Because $f$ is even : $f(x) =f(-x)$

letting $t = -x, dt = -dx$

$\int_{x=-a}^0 f(-x) \, dx = -\int_{u=a}^0 f(t) \, dt$=$\int_{u=0}^a f(t)dt$

Because $t$ is a dummy variable it can be replaced by x.

Is this rigorous proof ( is there any flaws? ) or is there better way of proving this?

Thanks in advance.

Kasmir Khaan
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1 Answers1

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It might be nit picking but your proof should be as follows:

Assume $f$ is odd. First observe \begin{align} \int^a_{-a}f(x)\ dx = \int^a_0 f(x)\ dx + \int^{0}_{-a} f(x)\ dx. \end{align} Using $x=-u$ substitution in the second integral yields \begin{align} \int^0_{-a} f(x)\ dx = -\int^0_{a} f(-u)\ du= \int^a_0 f(-u)\ du. \end{align} And since $f$ is odd, then it follows \begin{align} \int^a_0 f(-u)\ du = -\int^a_0 f(u)\ du \end{align} which means \begin{align} \int^a_{-a} f(x)\ dx = \int^a_0 f(x)\ dx -\int^a_0 f(u)\ du = 0. \end{align}

Jacky Chong
  • 25,739