Can anyone explain why $\int_{-\infty}^0 \frac{dx}{3-4x}$ does not exist? I know this will get $$-0.25 \lim_{t\to -\infty}(\ln (3-0)-\ln(3-4t))$$ As I know $$\ln(3+((-4)(-\infty)))=\ln(3+\infty)=\infty$$
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7WHat is your question? You seem to know already that $\int_{a}^{0}\frac{1}{3-4x},dx$ diverges to infinity as $a\to-\infty$. By definition, that's why the improper integral does not exist. – hmakholm left over Monica Dec 29 '13 at 22:12
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Why dose not exist why the answer is not ∞ ? – user32104 Dec 30 '13 at 06:12
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2In the common usage, "is infinity" describes one of several ways for a limit or improper integral to "not exist". – hmakholm left over Monica Dec 30 '13 at 11:34
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You have $${1\over 3 - 4x}\sim -{1\over4x}$$ as $x\to-\infty$. This last quantity does not integrate [finitely] at $-\infty$.
ncmathsadist
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Make the substitution $x=-t$, and you arrive at harmonic series-like sum, so it diverges.
Emolga
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I know why it's diverge but why it dose not exist because the answer should be ∞ ,right? – user32104 Dec 30 '13 at 06:15