Final answer:
The integral ∫[₁,[infinity])₁/ₓ²₁ dμ is evaluated by calculating a Lebesgue integral over [1, infinity) and adding the contribution of a Dirac delta measure at x=5. The result is 1 + 2*(1/25) = 27/25.
Step-by-step explanation:
The question asks to evaluate an integral with respect to the measure μ which is a sum of the Dirac delta measure at 5 and twice the Lebesgue measure. The integral to be evaluated is ∫[₁,[infinity])₁/ₓ²₁ dμ, which means we want to integrate the function 1/x^2 over the interval [1, infinity) with respect to the measure μ. Since the Dirac delta measure at 5 only affects the integral at the point x=5, the main part of the integral just involves integrating with respect to the Lebesgue measure.
To find the integral, we simply compute the Lebesgue integral of 1/x^2 from 1 to infinity, which is ∫[₁,∞) 1/x^2 dλ, and then add twice its value, according to the measure μ. The Dirac delta measure at 5 contributes a term (1/5^2) because the function under the integral is 1/x^2 and we evaluate it at x=5. Therefore, the evaluation of the integral is:
- Lebesgue integral part: ∫[₁,∞) 1/x^2 dλ = 1
- Delta measure at 5: (1/5^2)
Finally, the result is 1 (from the Lebesgue integral) + 2*(1/5^2) from the Dirac delta at 5, since μ includes twice the Lebesgue measure.
The complete evaluation of the integral is 1 + 2*(1/25) = 1 + 2/25 = 27/25.