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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?

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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.

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  • $\begingroup$ How do I find the derivative of the second integral without integrating? $\endgroup$ – Ruben Kruepper Aug 12 at 8:42
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    $\begingroup$ @RubenKruepper: Look up “Fundamental theorem of calculus” $\endgroup$ – Martin R Aug 12 at 8:49
  • $\begingroup$ @MartinR Doesn't the fact that our upper bound grows quadratically, not linearly, matter? $\endgroup$ – Ruben Kruepper Aug 12 at 8:53
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    $\begingroup$ @RubenKruepper: Sure (which means that you need the chain rule as well). $\endgroup$ – Martin R Aug 12 at 8:54
  • $\begingroup$ @MartinR I honestly thought of that two seconds after posting, thanks! $\endgroup$ – Ruben Kruepper Aug 12 at 8:55
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$$ let\; y(x)\;=\;\int_{1}^{x^2} \frac{\ln(xt)}{t+1}dt$$ Therefore y(x) = $ \int_{1}^{x^2} \frac{\ln(x)}{t+1}dt\; $ + $ \int_{1}^{x^2} \frac{\ln(t)}{t+1}dt $ $$ $$Hence y(x) = $\int_{1}^{x^2} \frac{\ln(t)}{t+1}dt\; $ + $ ln(x)\, \int_{1}^{x^2} \frac{1}{t+1}dt $ $$ $$ Now differentiate $$ \frac{dy}{dx}\;= \frac{2xlnx^2}{1+x^2}\; + \; lnx\,\frac{2x}{1+x^2}\;+\; (\frac{1}{x})(\; \int_{1}^{x^2}\frac{1}{1+t})dt $$ $$ $$ since all terms are positive for x $\gt1\;$hence f'(x)$\gt0$

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