If I integrate $\int_1^{-1} x^{-2} \,dx$ using normal integration methods (so without limits) the answer is $-2$. What is the significance of this answer? (Like what does it mean?) Thanks in advance
If I integrate $\int_1^{-1} x^{-2} \,dx =-2$ what is the significance of this answer?
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$\begingroup$
calculus
definite-integrals
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4The significance is it's wrong. – 2017-02-21
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0but for $x=0$ there is a pol for $$y=x^{-2}$$ – 2017-02-21
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0make sense using hadamard finite part method :) – 2017-02-21
2 Answers
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It doesn't mean anything. The integral does not converge there so the theorem that allows you to compute an integral as the difference of the anti derivatives does not apply and the result you get is meaningless.
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By definition: $\int_{-1}^{1} x^{-2} \,dx$ is convergent iff $\int_{-1}^{0} x^{-2} \,dx$ and $\int_{0}^{1} x^{-2} \,dx$ are convergent.
But $\int_{-1}^{0} x^{-2} \,dx$ and $\int_{0}^{1} x^{-2} \,dx$ are divergent.