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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!

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Check this theorem. – Mhenni Benghorbal May 3 '13 at 1:08
Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities. – Drake May 6 '13 at 13:51

1 Answer 1

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 $\int_{- \pi/4}^{\pi/4} f(x)dg(x)= \int_{- \pi/4}^{\pi/4}\frac{\sin^4x}{\cos^2x}d(1+\sin x)=\int_{- \pi/4}^{\pi/4}\frac{\sin^4x}{\cos^2x}\cos x dx$

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