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I'm having trouble solving this Riemann-Stieltjes integral:

$$\int_{- \pi/4}^{\pi/4} f(x)dg(x),$$ where $$f(x):= \begin{cases} \frac{\sin^4x}{\cos^2x}{} &\text{if }x\ge0, \\{}\\ \frac1{\cos^3x} &\text{if }x<0,\end{cases}$$

and $$g(x)=\begin{cases} \phantom{-} 1+\sin(x) &\text{if }-\pi/4 <x<\pi/4, \\ -1 &\text{otherwise}.\end{cases}$$

I believe the only jump discontinuities are at $-\pi/4$ and $\pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!

Batominovski
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Drake
  • 851

3 Answers3

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Let $T = \pi/4$. The first term in the Riemann-Stieltjes sum, $f(\xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 \downarrow -T$ and $\xi_1 \downarrow -T$, and similarly for the last term. The integral is $$I = \int_{-T}^T f(x) dg(x) = \\ \lim_{\epsilon \downarrow 0} \{ f(-T) (g(-T + \epsilon)- g(-T)) + \\ f(T) (g(T)- g(T - \epsilon)) \} + \\ \int_{-T}^T f(x) \cos x \,dx = \\ 2 \operatorname{arctanh}(\sqrt 2 - 1) + \frac {19 \sqrt 2} 6 - 2.$$ It can be verified that $$I = f(T) g(T) - f(-T) g(-T) - \int_{-T}^T g(x) df(x) = \\ f(T) g(T) - f(-T) g(-T) - \\ \lim_{\epsilon \downarrow 0} g(0) (f(\epsilon) - f(-\epsilon)) - \\ \int_{-T}^0 g(x) \left( \frac 1 {\cos^3 x} \right)' dx - \int_0^T g(x) \left( \frac {\sin^4 x} {\cos^2 x} \right)' dx.$$

Maxim
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Since $g$ is differentiable on $-\pi/4 <x<\pi/4$ your integral changes to Riemann integral simply by the following theorem: $$\int_a^bf(x)dg(x)=\int_a^bf(x)g'(x)dx$$ so you will have $\int_{- \pi/4}^{\pi/4} f(x)dg(x)= \int_{- \pi/4}^{\pi/4}\frac{\sin^4x}{\cos^2x}d(1+\sin x)=\int_{- \pi/4}^{\pi/4}\frac{\sin^4x}{\cos^2x}\cos x dx$

Ali
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  • The integral should really be $$\int_{-\pi/4}^{\pi/4} f(x) dg(x) = \int_{-\pi/4}^0 \frac 1{\cos^3 x} d(1+\sin x) + \int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(1+\sin x)$$ or $$\int_{-\pi/4}^{\pi/4} f(x) dg(x) = \int_{-\pi/4}^0 \frac 1{\cos^3 x} \cos x dx + \int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} \cos x dx$$ – New day rising Jan 07 '17 at 03:50
  • Psst... both of you need to take a closer look to what happens at the endpoints. – user361424 Jun 28 '17 at 00:55
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Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'\cdot f$

Namely from here we have $$\color{blue}{\int_a^bf(x)dg(x)=\int_a^bf(x)g'(x)dx}$$ Therefore, $$\int_{-\pi/4}^{\pi/4} f(x) dg(x) = \int_{-\pi/4}^0 \frac 1{\cos^3 x} d(1+\sin x) + \int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(1+\sin x)\\=\int_{-\pi/4}^0 \frac 1{\cos^2 x} dx + \int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(\sin x)$$

But $$\int_{-\pi/4}^0 \frac 1{\cos^2 x} dx =\int^{\pi/4}_0 (\tan x)' dx =1$$

and $$\int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(\sin x) = \int_0^{\pi/4} \frac{\sin^4 x}{1-\sin^2 x} d(\sin x) =\int_0^{\sqrt{2}/2} \frac{t^4 }{1-t^2 } dt\\=\int_0^{\sqrt{2}/2} -t^2+\frac{t^2 }{1-t^2 } dt$$ $$ $$

Guy Fsone
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  • The first formula isn't applicable when $g$ has jump discontinuities. – Maxim May 28 '18 at 21:15
  • @Maxim It is possible the discontinuities are of first order. – Guy Fsone May 30 '18 at 07:14
  • There are zero order discontinuities at the endpoints. See my answer. – Maxim May 30 '18 at 09:42
  • @Maxim what do you mean by zero order discontnuity? – Guy Fsone May 31 '18 at 01:52
  • I mean that the function itself (the derivative of zero order) is discontinuous. – Maxim May 31 '18 at 09:45
  • discontinuous mean non differentiable. – Guy Fsone Jun 02 '18 at 07:27
  • A discontinuous function is said to have a zero order discontinuity because, formally, we can say that the function is its own derivative of order zero. If your definition of the order of discontinuities is different, please state it. The upshot is that your answer is not correct because $g$ is not continuous from the right at the left endpoint and not continuous from the left at the right endpoint. You forgot to include the contributions from those discontinuities. – Maxim Jun 02 '18 at 12:27