# Convergence of integrals over divergent parts

I'm wondering if it is possible for an integral which diverges in the limits $1$ to $\infty$ to converge in the limits from $0$ to $\infty$. And if so: how could I find this out?

For example $$\int{e^x*(x-1)\over x^2}dx = \frac{e^x}{x}$$ would diverge in $[\epsilon,\infty)$ to $\infty$ and diverge from $[\epsilon,0)$ to $-\infty$, which would mean that the integral in $(0,\infty)$ is $\infty-\infty$

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Check this out: VP –  Ilya Jan 29 '13 at 12:32
Under the usual calculus definition of convergence of an improper integral, the answer is no. If it is "bad" over $[1,\infty)$, then it is bad over $[0,\infty)$. In general, the "difference" of "infinities" does not make sense. –  André Nicolas Jan 29 '13 at 12:37
Most likely you think of $\int_a^b + \int_b^c = \int_a^c$ here. Beware that this equation only holds if all integrals exist. –  mkl Jan 29 '13 at 12:42
If the integral diverges over $[0, \infty)$, it could converge over $(-\infty, \infty)$. Isn't that because of subtraction of two infinity-values? So what's the difference here? –  user1329846 Jan 29 '13 at 12:57