In doing some old exam questions, I came across the following problem. Let $f:(1,\infty)\rightarrow \mathbb{R}$
$$f(x)=\int_{1}^{x^2} \frac{\ln(xt)}{1+t}dt$$
Questions:
a) Reason as to why f is continuously differentiable and find an integral-free representation of the derivative
b) Show that $f'(x)>0$ for $x>1$
I assume (b) will be trivial when (a) is solved.
For (a) my approach was integrating more or less directly, then differentiating. I end up caught in a cycle of integrating by parts:
$$\int_{1}^{x^2} \frac{\ln(1+t)}{t}dt$$and
$$\int_{1}^{x^2} \frac{\ln(t)}{t+1}dt$$
keep coming back… I don't have anything in my toolbox (that I know of) that lets me crack this.
So I assume that either I shouldn't actually integrate and then differentiate or I'm missing something in my "integration-toolbelt". What's going on here? Also, I don't actually know how to "Reason as to why f is continuously differentiable", are my problems connected?
Best Answer
Hint: Use the fact that $\ln(xt)=\ln x+\ln t$. So $f(x)=\ln \, x\int_1^{x^{2}} \frac 1 {1+t} dt+\int_1^{x^{2}} \frac {\ln t } {1+t}dt$ and the first term is $(\ln x) (\ln (1+x^{2})-\ln 2)$. You can now write down $f'$ easily.
It is also quite easy to show that $f'(x) >0$ for $x >1$. I will leave that to you.