Prove that there does not exist an elemenary function $f$ such that $f({\rm glog}x)$ is an antiderivative of ${\rm glog}x$.
(${\rm glog}x$ denotes the inverse function of $e^x/x$ and is called generalized logarithm.)
Solution
Suppose , on the contrary , that such function exists that is:
$$\frac{\mathrm{d} }{\mathrm{d} x}\left ( f\left ( {\rm glog}x \right ) \right )={\rm glog}x$$
Therefore:
$$\begin{aligned}\frac{\mathrm{d} }{\mathrm{d} x}\left ( f\left ( {\rm glog}x \right ) \right )={\rm glog}x &\Leftrightarrow f' \left ( {\rm glog}x \right ){\rm glog}'x= {\rm glog}x\\ &\Leftrightarrow f'\left ( {\rm glog}x \right )\cdot \frac{{\rm glog}x}{x {\rm glog}x-x}= {\rm glog}x\\ &\Leftrightarrow f'\left ( {\rm glog}x \right )= x {\rm glog}x -x \end{aligned}$$
However by subbing $x=e^y/y$ we get that:
$$ f'(y)=e^y -\frac{e^y}{y}$$
leading us to the fact that $e^y/y$ is an elementary integrable function , which is impossible since we know that it does not have an elementary antiderivative. So, the exercise comes to an end!
(${\rm glog}x$ denotes the inverse function of $e^x/x$ and is called generalized logarithm.)
Solution
Suppose , on the contrary , that such function exists that is:
$$\frac{\mathrm{d} }{\mathrm{d} x}\left ( f\left ( {\rm glog}x \right ) \right )={\rm glog}x$$
Therefore:
$$\begin{aligned}\frac{\mathrm{d} }{\mathrm{d} x}\left ( f\left ( {\rm glog}x \right ) \right )={\rm glog}x &\Leftrightarrow f' \left ( {\rm glog}x \right ){\rm glog}'x= {\rm glog}x\\ &\Leftrightarrow f'\left ( {\rm glog}x \right )\cdot \frac{{\rm glog}x}{x {\rm glog}x-x}= {\rm glog}x\\ &\Leftrightarrow f'\left ( {\rm glog}x \right )= x {\rm glog}x -x \end{aligned}$$
However by subbing $x=e^y/y$ we get that:
$$ f'(y)=e^y -\frac{e^y}{y}$$
leading us to the fact that $e^y/y$ is an elementary integrable function , which is impossible since we know that it does not have an elementary antiderivative. So, the exercise comes to an end!
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