Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{0} \frac{x}{x^{2}+1} d x $$

Short answer

The integral converges and equals to \(π/2\).

Step-by-step solution

Substitution

First, instead of directly dealing with -∞, let’s put a variable b in place of -∞. Now, the integral becomes \(\int_{b}^{0} \frac{x}{x^{2}+1} dx\). At the end, we’ll take the limit of b approaching -∞.

Evaluate the Integral

To get the integral, one could recognize that the integrand is the derivative of \(arctan(x)\), hence \(\int_{b}^{0} \frac{x}{x^{2}+1} dx = [arctan(x)]_{b}^0\).

Find the limit as b approaches -∞

After finding the antiderivative of the integrand, substitute the bounds of the integral back in: \(= arctan(0)-arctan(b)\). Now, take the limit of this as \(b\) approaches \(-∞\). This becomes: \(\lim_{b \to -\infty} [arctan(0) - arctan(b)] = 0 - -π/2 = π/2\).