For an odd function $f: \mathbb R \rightarrow \mathbb R $ we have $$\int_{-a}^{a} f(x) dx = 0$$ and for an even function $f: \mathbb R \rightarrow \mathbb R $ we have $$\int_{-a}^{a} f(x) dx =2\int_{0}^{a} f(x) dx$$ Using these two results I'm trying to find the necessary conditions, for odd $f$ and even $f$, such that the integral $\int_{-\infty}^{\infty} f(x) dx$ converges.
For an even function $f: \mathbb R \rightarrow \mathbb R $, $\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty} f(x) dx$. So, $\int_{-\infty}^{\infty} f(x) dx$ converges if $\int_{-\infty}^{0} f(x) dx$ and $\int_{0}^{\infty} f(x) dx$ both converge. We can extend this by saying $\int_{0}^{a} f(x) dx$ has to converge as $a \rightarrow \infty$ (since $\int_{-a}^{0} f(x) dx = \int_{0}^{a} f(x) dx $ for even functions). Here I haven't used the result above directly because I know that it's not valid to say $\int_{-\infty}^{\infty} f(x) dx=\int_{-a}^{a} f(x) dx$ as $a \rightarrow \infty$. So I'm wondering how I can use the above result directly to find the condition for te even case.
For an odd function $f: \mathbb R \rightarrow \mathbb R $, $\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty} f(x) dx$. So, $\int_{-\infty}^{\infty} f(x) dx$ converges if $\int_{-\infty}^{0} f(x) dx$ and $\int_{0}^{\infty} f(x) dx$ both converge. We can extend this by saying $\int_{0}^{a} f(x) dx$ has to converge as $a \rightarrow \infty$ (since $\int_{-a}^{0} f(x) dx = -\int_{0}^{a} f(x) dx $ for odd functions). Like for the even case, I'm not sure how I can use the earlier results directly to find the condition for the odd case.
hint
by definition
$$\int_{-\infty}^{\infty}f (t)dt $$ converges
$$\iff $$ $$\lim_{(x,y)\to (-\infty,+\infty)}\int_x^yf (t)dt $$ exists in $\Bbb R $.
$$\lim_{a\to +\infty}\int_{-a }^axdx=0$$ but $$\int_{-\infty}^{+\infty}xdx $$ diverges.