Find Value of $\int_0^1 (f(x)-g(x))dx$ with $f,g$ Continuous

[Blandongstein]

Let $f$ be a continuous function for $x \in (0,1]$ and $\displaystyle g(x)=\int_{1}^{1 \over x}\frac{1}{t}f\left( \frac{1}{t}\right)dt$, then find the value of

$$ \int_0^1 (f(x)-g(x))dx$$

 

[sbhatnagar]

If $\displaystyle g(x) = \int_{1}^{\frac{1}{x}}\frac{1}{t}f\left( \frac{1}{t}\right)dt$ then

$\displaystyle g'(x)=-\frac{xf(x)}{x^2} =-\frac{f(x)}{x}$.

or $f(x)=-xg(x)$ ...(1)

$\displaystyle I = \int_{0}^{1}(f(x)-g(x))dx = \int_0^1f(x)dx-\int_0^1g(x)dx$

Use integration by parts on the second integral:

$\displaystyle I = \int_0^1f(x)dx -(xg(x))_0^1 + \int_0^1 xg'(x) dx$

by (1) we have
$\displaystyle I= \int_0^1f(x)dx - \int_0^1f(x)dx -g(1) = -g(1) = -\int_{1}^{1}\frac{1}{t}f\left( \frac{1}{t}\right)dt $

$=0$

 

[MarkFL]

Nearly identical to the method I had in mind. :D