Prove that if $$\int^{\infty}_{-\infty}f(t)dt$$converges then it is equal to $$\lim_{c \to \infty}\int^c_{-c}f(t)dt.$$ $$$$As we know that $$\begin{aligned}\int^{\infty}_{-\infty}f(t)dt&=\int^a_{-\infty}f(t)dt+\int_a^{\infty}f(t)dt\\&=\lim_{c \to \infty}\int^a_{-c}f(t)dt+\lim_{d \to \infty}\int_a^df(t)dt\end{aligned}$$ and as it is given that $\int^{\infty}_{-\infty}f(t)dt$ converges, so both the limits on the RHS of the above equation exists. Now as $$\lim_{d \to \infty}\int_a^df(t)dt$$ exists so it is equal to $$\lim_{c \to \infty}\int_a^cf(t)dt$$ and hence we get $$\begin{aligned}\int^{\infty}_{-\infty}f(t)dt&=\int^a_{-\infty}f(t)dt+\int_a^{\infty}f(t)dt\\&=\lim_{c \to \infty}\int_{-c}^af(t)dt+\lim_{c \to \infty}\int_a^cf(t)dt\\&=\lim_{c \to \infty}\int_{-c}^cf(t)dt\end{aligned}.$$ $$$$Is My Proof Correct??
2 Answers
The value of $\lim_{c \to \infty}\int^c_{-c}f(t)\,dt$ (when it exists) is called the Cauchy principle value. I would instead write your proof as follows:
Assume that $\int_{-\infty}^b f(t)\,dt$ and $\int_b^\infty f(t)\,dt$ exist. Then $$I=\int_{-\infty}^\infty f(t)\,dt = \lim\limits_{a \to -\infty} \int_a^b f(t)\,dt + \lim\limits_{c \to \infty} \int_b^c f(t)\,dt = \lim\limits_{c \to \infty} \int_{-c}^b f(t)\,dt + \lim\limits_{c \to \infty} \int_b^c f(t)\,dt.$$
As the two limits exist, we can combine them to get
$$I=\lim\limits_{c \to \infty} \left(\int_{-c}^b f(t)\,dt + \int_b^c f(t)\,dt \right) =\lim\limits_{c \to \infty} \int_{-c}^c f(t)\,dt,$$
which is the same as the Cauchy principle value. The Cauchy principle value and improper integral are only the same when the integral converges.
As an example, consider $\int_{-\infty}^\infty t\,dt$. Observe that $\int_0^\infty t\,dt$ diverges to $\infty$ and $\int_{-\infty}^0 t\,dt$ diverges to $-\infty$. Therefore, the integral diverges from either side. However, $$\lim\limits_{c \to \infty} \int_{-c}^c t\,dt = \lim\limits_{c \to\infty} \left(\frac{c^2}{2} - \frac{(-c)^2}{2}\right) = 0.$$
Thus the principal value of $\int_{-\infty}^\infty t\,dt$ is zero while the integral itself diverges. They will be the same if and only if the integral converges.
No! $$\int^{\infty}_{-\infty}f(t)dt=\lim_{c \to \infty}\int^{-c}_{-\infty}f(t)dt+\lim_{c \to \infty}\int^{+c}_{-c}f(t)dt+\lim_{c \to \infty}\int^{+\infty}_{+c}f(t)dt\tag{1}$$ You used $$\lim_{c \to \infty}\int^{+\infty}_{+c}f(t)dt=\lim_{c \to \infty}\int^{-c}_{-\infty}f(t)dt=0$$ which is what is to be proven by the logic of convergence. After using the above, you play with the middle term of $(1)$ anyhow, but the proof is left way behind.
