Sign of an improper integral

  • #1
18
1
Let ##f:[0;1)\to\mathbb{R}## and ##f\in C^1([0;1))## and ##\lim_{x\to1^-}f(x)=+\infty## and ##\forall_{x\in[0;1)}-\infty<f(x)<+\infty##. Define $$A:=\int_0^1f(x)\, dx\,.$$ Assuming ##A## exists and is finite, is it possible that ##\text{sgn}(A)=-1##?
 
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Answers and Replies

  • #2
f(x) < 0 for much of the interval.
 
  • #3
Let ##f:[0;1)\to\mathbb{R}## and ##f\in C^1([0;1))## and ##\lim_{x\to1^-}f(x)=+\infty## and ##\forall_{x\in[0;1)}-\infty<f(x)<+\infty##. Define $$A:=\int_0^1f(x)\, dx\,.$$ Assuming ##A## exists and is finite, is it possible that ##\text{sgn}(A)=-1##?
Try $$f(x)=\frac{1}{\sqrt{1-x}}-3.$$
 

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