Today I learned a bit about integrals and I'd like to know if that is true, or if there is a counterexample that disproves the following:

When an $(\alpha\in \mathbb{R})$ exists, so that for all sequences $((a_k)_{k\in \mathbb{N}}, (b_k)_{k\in \mathbb{N}} \subset\mathbb{R})$ with $(a_k\to -\infty, b_k\to \infty)$ for $(k\to \infty)$ already $(\lim_{k\to \infty} \int_{a_k}^{b_k} f(x)\,\mathrm{d}x=\alpha)$ applies, then the following $(\alpha=\int_{-\infty}^{\infty} f(x)\, \mathrm{d}x)$ applies too.

My thoughts are that when $(a_k\to -\infty, b_k\to \infty)$ for $(k\to \infty)$ applies, then at some time (meaning for $(k\geq k_0)$ with a $(k_0\in \mathbb{N}))$ the following: $(a_k<0)$ and $(b_k>0)$ hold.

So the integral can be split like that: $(\int_{a_k}^{b_k} f(x)\mathrm{d}x=\int_{a_k}^{0} f(x)\mathrm{d}x+\int_{0}^{b_k} f(x)\mathrm{d}x)$ which leads us to the definition of improper integrals.

Is my argumentation correct or wrong? Because from $(\lim_{n\to \infty} (c_n+d_n))$ the existence of $(\lim_{n\to \infty}c_n)$ or $(\lim_{n\to \infty}d_n))$ do not follow...


The claim is correct: Using the assumption show that


for any two sequences $c_k,d_k\rightarrow\infty$, or equivalently


Hence By Cauchy $\lim_{x\rightarrow\infty}\int_0^xf$ converges. Similarly $\lim_{x\rightarrow-\infty}\int_x^0f$ converges so overall $f$ is integrable over $\mathbb{R}$. Choosing $a_k=-k,\ b_k=k$ gives you the integral is $\alpha$.

Your argument is wrong for the reason you have mentioned.


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