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(Note: I asked this question a few days ago on math.stackexchange but didn't get any responsesand I'm . I've therefore postit decided to post it here instead.)

I have a problem justifying the throwing away the divergent term in order to obtain Hadamard's finite part. I find this step to be highly unusual and it is not obvious to me how the resulting expression can be valid. I'd appreciate, if possible, intuitive arguments -- I assume there are some. Its not the mathematics that I have problems with at this stage. Its the intuition.

For example (taken from the Wikipedia page) the finite part of the following integral $$ \int_a^b \frac{f(t)}{(t-x)^2}\, dt = \lim_{\varepsilon \to 0} \left[ \int_a^{x-\varepsilon}\frac{f(t)}{(t-x)^2}\,dt + \int_{x+\varepsilon}^b\frac{f(t)}{(t-x)^2}\,dt -\frac{2f(x)}{\varepsilon} \right] $$

involves throwing away the term $\frac{2f(x)}{\varepsilon}$. I find it hard to justify this step especially when the term is neither finite nor negligible.
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Rationale for Hadamard's finite part of a divergent integral

(Note: I asked this question a few days ago on math.stackexchange but didn't get any responses and I'm therefore postit it here instead.)

I have a problem justifying the throwing away the divergent term in order to obtain Hadamard's finite part. I find this step to be highly unusual and it is not obvious to me how the resulting expression can be valid. I'd appreciate, if possible, intuitive arguments -- I assume there are some. Its not the mathematics that I have problems with at this stage. Its the intuition.

For example (taken from the Wikipedia page) the finite part of the following integral $$ \int_a^b \frac{f(t)}{(t-x)^2}\, dt = \lim_{\varepsilon \to 0} \left[ \int_a^{x-\varepsilon}\frac{f(t)}{(t-x)^2}\,dt + \int_{x+\varepsilon}^b\frac{f(t)}{(t-x)^2}\,dt -\frac{2f(x)}{\varepsilon} \right] $$

involves throwing away the term $\frac{2f(x)}{\varepsilon}$. I find it hard to justify this step especially when the term is neither finite nor negligible.