Question

Determine if the improper integral converges or diverges. if it converges, what does it converge to?

∫[infinity]-[infinity] x/x²+1 dx

Answer 1

The improper integral
`∫[-∞ to ∞] x/(x² + 1) dx` converges, and it converges to 0.

Step-by-step explanation:

The given integral is
`∫[-∞ to ∞] x/(x² + 1) dx`. To determine convergence, we can split the integral into two parts:
`∫[-∞ to 0] x/(x² + 1) dx and ∫[0 to ∞] x/(x² + 1) dx`, and then evaluate each separately.

For the first part,
`∫[-∞ to 0] x/(x² + 1) dx`, we can make a substitution, letting
`u = x² + 1`. Then,
`du = 2x dx`. The integral becomes
`(1/2) ∫[-∞ to 1] du/u`, which is the natural logarithm evaluated from 1 to ∞. As the natural logarithm of ∞ is infinite, this part diverges.

For the second part,
`∫[0 to ∞] x/(x² + 1) dx`, we use a similar substitution, letting
`u = x² + 1`. Then,
`du = 2x dx`. The integral becomes
`(1/2) ∫[1 to ∞] du/u`, which is the natural logarithm evaluated from 1 to ∞. As the natural logarithm of ∞ is infinite, this part also diverges.

However, when we combine the two parts, the divergent infinities cancel each other out, resulting in a convergent integral that converges to 0. This cancellation is a common occurrence in integrals with odd symmetry, where the areas above and below the x-axis offset each other, leading to convergence despite individual divergences.