I need help with this problem.
Let f(x) = int_{1,x} (log t)/(t + 1) dt if x > 0. Compute f(x) + f(1/x).
Thanks
I'm thinking the second fundamental theorem of calculus, but what you have is not exactly in the form to use it. to apply the second fundamental theorem of calculus, we should have something like:
$\displaystyle f(x) = \frac {d}{dt} \int_1^x \frac {log(t)}{t + 1} dt$
But you don't have the $\displaystyle \frac {d}{dt}$
So i guess we should actually find the integral and evaluate it between those limits to find f(x) and hence f(1/x). But i don't want to waste my time doing this if we don't have to, so:
(1) Check if there is a typo here and that we should actually have a $\displaystyle \frac {d}{dt} $ somewhere, or
(2) If we can still apply the second fundamental theorem of calculus, could someone please confirm that?
$\displaystyle f(x) = \int_1^x \frac{\ln t}{1+t} dt$
$\displaystyle f(1/x) = \int_1^{1/x} \frac{\ln t}{1+t} dt$
Note, $\displaystyle \frac{\ln t}{1+t} = \frac{ - \ln \frac{1}{t} }{t^2+t} = \frac{-\ln \frac{1}{t}}{1} \cdot \frac{1}{1+\frac{1}{t}} \cdot \frac{1}{t^2}$
In the second integral define $\displaystyle s=\frac{1}{t}$ then, $\displaystyle s=-\frac{1}{t^2}$.
Henceforth,
$\displaystyle \int_1^{1/x} \frac{\ln t}{1+t} dt = \int_1^{1/x} \frac{-\ln \frac{1}{t}}{1} \cdot \frac{1}{1+\frac{1}{t}} \cdot \frac{1}{t^2} dt$
Substitution theorem with $\displaystyle s=\frac{1}{t}$:
$\displaystyle \int_1^s \frac{\ln s}{1+s} ds = f(x)$
Thus,
$\displaystyle f(1/x) = f(x)$.