What assumptions are needed for $\int_{-a}^{a} f(x)\,\mathrm{d}x=\int_{0}^{a} f(x)+f(-x)\,\mathrm{d}x$?

Question

This might seem like a strange question, but I was wondering what assumptions is needed for

$\displaystyle \hspace{1cm}\int_{-a}^{a} f(x)\,\mathrm{d}x=\int_{0}^{a} f(x)+f(-x)\,\mathrm{d}x$

I am not asking for under which assumptions the integral exists/converges only, under which assumptions the equality holds.

Obviously for the integral to converge we need $f$ to be integrable in some form, however is that needed just for the equality to hold.

Similarly does $a$ need to be real, what restrictions is needed upon $a$?

Answer 1

I will assume that you are talking about Riemann integrals.

• If $a \in \mathbb{R}_+$ and $f$ is Riemann-integrable on $[-a,a]$, then the equality holds, as user289143 showed it.

• If $f$ is not Riemann-intehrable, then the LHS is meaningless, but the RHS can still be zero, because if $f$ is odd, then $f(x)+f(-x)=0$, and it's integrable.

• If $f$ is improper Riemann-integrable on $\mathbb{R}$, and $a=\infty$, the equality still holds, you just need to take the limit as $a \to +\infty$

• If $f$ is not improper Riemann-integrable on $\mathbb{R}$, the LHS is meaningless, but the RHS can be zero, if $f$ is odd.

• If $f$ is unbounded, improper Riemann-integrable on $[-a, a]$ for $a \in \mathbb{R}_+$, then the equality still holds. We can assume that there is only one singularity on $[-a, a]$, because if there were more, we would be able to split it into smaller intervals, with only $1$ singularity in each. So let's assume that the singularity is at $b \in (0, a)$ (the edge-case, i.e. $b=a$ is the same, with only one limit). Then the LHS is, by definition: $$\lim_{q \to b-0}\int_{-a}^b f + \lim_{p \to b+0} \int_b^a f$$ And the RHS is: $$\lim_{q \to b-0}\int_{0}^b f(x)+f(-x) \mathrm{d}x + \lim_{p \to b+0} \int_b^a f(x) + f(-x) \mathrm{d}x$$ And you can just split up the integral in the LHS to the intervals $[-a, 0]$ and $[0, q]$, because the limit exists, and pull out the $[-a, 0]$ part. In the RHS, you can use the additivity of the integral to separate the integral of $f(-x)$ (and the existence of the limits), and after a substitution, note that the two sides are the same.

• If $f$ is unbounded, not improper Riemann-integrable on $[-a, a]$ for $a \in \mathbb{R}_+$, then the LHS is meaningless, but the RHS can be zero, if $f$ is odd.

Answer 2

$\int_{-a}^af(x)dx=\int_{-a}^0 f(x)dx+\int_0^a f(x)dx$, so you're asking when $$ \int_{-a}^0f(x)dx=\int_0^af(-x)dx $$ and this is trivially true because with the substitution $u=-x$, we have $$ \int_{-a}^0f(x)dx=\int_{a}^0f(u)du=\int_{a}^0f(-x)(-dx)=-\int_a^0f(-x)dx=\int_0^af(-x)dx $$