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## Main Question or Discussion Point

Hi,

I have an integral from 0 to ∞ and divided into two integrals, one from 0 to C and one from C to ∞.

I'm trying to simplify the second integral so that I can solve for C to make the integral less than ε.

[itex]\int\frac{1-e^{-x^{3}}}{x^{2}} = -xf(x) + \int 3xe^{-x^{3}} < \int 3xe^{-x} = -3xe^{-x} + \int 3e^{-x} < \int 3e^{-x} = -3e^{-x} < ε[/itex]

For ε = 1e^-6 and limits from C to ∞ this gives me: [itex]3e^{-C} < 10^{-6} \Rightarrow C ≥ ln(3) + 6ln(10) ≈ 15[/itex].

However, Wolfram gives me that the value of the integral from 15 to ∞ is ≈0.0667 which is far from 1e^-6.

Is my inequality wrong?

I have an integral from 0 to ∞ and divided into two integrals, one from 0 to C and one from C to ∞.

I'm trying to simplify the second integral so that I can solve for C to make the integral less than ε.

[itex]\int\frac{1-e^{-x^{3}}}{x^{2}} = -xf(x) + \int 3xe^{-x^{3}} < \int 3xe^{-x} = -3xe^{-x} + \int 3e^{-x} < \int 3e^{-x} = -3e^{-x} < ε[/itex]

For ε = 1e^-6 and limits from C to ∞ this gives me: [itex]3e^{-C} < 10^{-6} \Rightarrow C ≥ ln(3) + 6ln(10) ≈ 15[/itex].

However, Wolfram gives me that the value of the integral from 15 to ∞ is ≈0.0667 which is far from 1e^-6.

Is my inequality wrong?