Solve Real-Life Problems w/ Stieltjes Integral

  • Thread starter Thread starter r4nd0m
  • Start date Start date
  • Tags Tags
    Integral
r4nd0m
Messages
96
Reaction score
1
Can you give me a simple real-life problem, where you need to use Stieltjes integral and can you show how you proceed in solving this kind of problems?
 
Physics news on Phys.org
I'm not certain what you consider "real life"! I suppose it wouldn't help for me to point out that in "real life" you have to take a calculus test.

The Stieljes integral differs from the ordinary Riemann integral in that, after we have divided the interval (a to b, say) into n intervals with endpoints xi, xi+1, instead of defining Delta x<sub>i</sub> to be simply x<sub>i+1</sub>- x<sub>i</sub>, that is, the length of the interval, we define it to be \alpha(x_{i+1})- \alpha(x_i) where \alpha(x) can be any increasing function. Taking the &quot;Riemann sums&quot; as usual then and taking the limit as the number of intervals goes to infinity results in the Stieltjes integral \int f(x)d\alpha rather than the Riemann integral \int f(x)dx.<br /> <br /> Of course if \alpha(x) happens to be differentiable then it is easy to see that<br /> \int f(x)d\alpha= \int f(x)\alpha&amp;#039;(x)dx<br /> <br /> One common application is this: let \alpha(x) be the &quot;step&quot; function (f(x)= 0 for 0&lt;= x&lt; 1, f(x)= 1 for 1&lt;= x&lt; 2, etc.). Then the sum <br /> \Sum_{n=0}^\k f(n)<br /> can be written as the Stieltjes integral<br /> \int_0^{n+1} f(x)d\alpha<br /> allowing one to combine the theory of sums with integrals.<br /> <br /> <br /> In particular
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top