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How do you compute the following? $$\int_{-\pi/4}^{\pi/4} f(x) \, dg(x) $$ Given $$f(x) = \begin{cases} \dfrac{\sin^4(x)}{\cos(x)} & x \in [0,\infty)\\[6pt] \dfrac{1}{\cos^3(x)} & x \in(-\infty,0) \end{cases} $$ Alright, so we have a discontinuity at $0$. Can I just evaluate as: $$\int_{-\pi/4}^{\pi/4} f(x) \, dg(x)=f(0)(g(0^{+})-g(0^{-1}))+\int_{-\pi/4}^{\pi/4} f(x)g'(x) \, dx$$ I get confused looking at examples, because the integral term always somehow goes to 0. I'm not sure if that's what happens in this case. The first term of the left hand side is $0$, so I think my answer here is $0$, but I'm not sure. Would appreciate an explanation for what happens to the integral term.

EDIT: $$g(x)=1+\sin(x)$$ for $$x \in (-\frac{\pi}{4},\frac{\pi}{4})$$ and $-\frac{1}{2}$ elsewhere

Taylor
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  • The fact that you have a discontinuity in $f$ seems to be no obstacle: this looks to me like an ordinary Riemann integral. – Ian May 06 '15 at 00:05
  • To clarify, there is some issue if and only if you have $a,b$ such that $g(b)-g(a) \neq \int_a^b g'(x) dx$. Otherwise you just integrate $f g'$ in the Riemann sense. – Ian May 06 '15 at 00:49
  • @Ian You need to take the endpoints into account as well, the condition that you wrote is violated if $a = -\pi/4$ or $b = \pi/4$. – Maxim May 28 '18 at 23:04

1 Answers1

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If you take $g(x)=1 +\sin(x)$ then $\int_{0}^{\pi/4} f(x)\cos(x)dx=\int_{0}^{\pi/4} \sin^4(x) = \frac{1}{32}(3\pi -8)$ and $\int_{-\pi/4}^{0} f(x)\cos(x)dx=\int_{-\pi/4}^{0} \frac{1}{\cos^2(x)} = 1$. So $\int_{-\pi/4}^{\pi/4} f(x) \, dg(x) = 1 + \frac{1}{32}(3\pi -8)$

GovernmentFX
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