Question

How to show that

{∣lns−1/2⋅ln(s 2 +4)∣} from 0 to [infinity] = ln((s^2+4)/s^2)?

a. Use the properties of logarithms.

b. Apply the rule of exponents.

c. Employ L'Hôpital's Rule.

d. None of these

Answer 1

To show that {∥ln(s-1/2·ln(s^2+4))∥} from 0 to ∞ is equal to ln((s^2+4)/s^2), we can use the properties of logarithms and integrate the expression.

Step-by-step explanation:

To show that {∥ln(s-1/2·ln(s^2+4))∥} from 0 to ∞ is equal to ln((s^2+4)/s^2), we can use the properties of logarithms.

1. First, let's simplify the expression inside the absolute value.

• ln(s-1/2) is equivalent to ln√s.
• ln(s^2+4) is unaffected by the absolute value, since it is always positive or zero.

Now, let's simplify the absolute value expression.

• ln√s is always positive, so the absolute value is unnecessary.
• ln(s^2+4) remains the same.

Finally, let's integrate the simplified expression.

• Integrating ln(s^2+4) with respect to s gives s^2+4.
• Integrating ln√s with respect to s gives 2ln√s.

Combining the integrals gives (s^2+4) - 2ln√s.Taking the limits from 0 to ∞ allows us to evaluate the expression.

• Substituting ∞ gives (∞^2+4) - 2ln∞.
• Substituting 0 gives (0^2+4) - 2ln0.

Since ln0 is undefined, the integral from 0 to ∞ does not converge.