I Is the sign of the integral of this function negative?

Rlwe
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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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f(x) < 0 for much of the interval.
 
Rlwe said:
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 $$f(x)=\frac{1}{\sqrt{1-x}}-3.$$
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
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