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
$$\lim_{k\rightarrow\infty}\int_{c_k}^{d_k}f=0$$
for any two sequences $c_k,d_k\rightarrow\infty$, or equivalently
$$\lim_{x,y\rightarrow\infty}\int_x^yf=0$$
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.