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How do you evaluate the integral #int xlnxdx# from 0 to #oo#?

Answer 1

Charles Arocha

You can't as the integral is divergent

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Answer 2

Aurora Alvarado

To evaluate the integral ∫xln(x)dx from 0 to ∞, we use integration by parts. Let u = ln(x) and dv = xdx. Then du = (1/x)dx and v = (1/2)x^2.

Applying the integration by parts formula:

∫u dv = uv - ∫v du

We get:

∫xln(x)dx = (1/2)x^2 ln(x) - ∫(1/2)x^2(1/x)dx = (1/2)x^2 ln(x) - (1/2)∫xdx = (1/2)x^2 ln(x) - (1/4)x^2

Now we evaluate this from 0 to ∞:

lim(a→∞) [(1/2)a^2 ln(a) - (1/4)a^2] - lim(a→0) [(1/2)a^2 ln(a) - (1/4)a^2]

We evaluate the limits separately. As a approaches infinity, ln(a) grows slower than any power of a, so the first term dominates, and the second term becomes negligible. Therefore, the first limit approaches infinity. The second limit is 0 since ln(a) approaches negative infinity as a approaches 0.

Thus, the integral diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.