Find the value of the function

Calculus Level pending

A function $f(x)$ satisfies the equation

$f'(x) =\ln (\ln x) +\dfrac {1}{(\ln x) ^2}$

and it's value at $x=e$ is $f(e)=0$ . Here $e$ is the Euler's Constant, $\approx 2.71828182846$ .

Find the value of $f(10)$ .

The answer is 6.716.

1 solution

Ron Gallagher
May 18, 2020

We need to find an anti-derivative of f'(x). To do this, make the substitution u = ln(x). This means x = exp(u) and dx = exp(u) du. The integrand then becomes (ln(u) + u^-2) exp(u) = ln(u) exp(u) + (u^2) exp(u). The integral of the first term can be integrated by parts twice to complete integration (there will be a "-(u^2) exp(u)" term that will cancel the second integral). Upon doing this, we find that exp(u) (ln(u) - 1/u) + C = x (ln(ln(x) - 1/ln(x)) + C = f(x) (here, C is a constant). But, f(e) = e (ln(1) - 1) + C = 0, so that C = e. Hence, f(x) = x (ln(ln(x) - 1/ln(x)) + e, so that f(10) = 6.716 (approximately).

Find the value of the function

The answer is 6.716.

1 solution

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