Question

Let $f:(0,\infty )\to (0,\infty )$ be a differentiable function satisfying, $x\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt=\underset{0}{\overset{x}{\int }}tf\left(t\right)dt,x\in R^{+}$ and $f\left(1\right)=1$, the value of $f\left(x\right)=\frac{1}{x^m}{e}^{(1-1/x)}$, then $m=$

Answer 1

$x\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt=\underset{0}{\overset{x}{\int }}tf\left(t\right)dt$
Differentiating w.r.t. $x$,
$x(1-x)f\left(x\right)+\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt=xf\left(x\right)\Rightarrow x^{2}f\left(x\right)=\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt$
Differentiating w.r.t. $x$,
$\Rightarrow x^2{f}^{\prime }\left(x\right)+2xf\left(x\right)=(1-x)f\left(x\right)\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1-3x}{x^2}\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int \frac{1-3x}{x^2}dx\Rightarrow \log f\left(x\right)=-\frac{1}{x}-3\log x+\log c$
We know that $f\left(1\right)=1$
$0=-1+\log c\Rightarrow c=e$
$\Rightarrow \log f\left(x\right)=-\frac{1}{x}-3\log x+1\Rightarrow f\left(x\right)=\frac{1}{x^3}{e}^{(1-1/x)}\Rightarrow m=3$