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Suppose $f$ is Riemann integrable on $[a,b]$ and $g$ be continuously differentiable on $[a,b]$.

Define $$F(x):= \int_a ^x f(t) dt,\ \ \forall x\in[a,b]$$

We know $F$ is continuous hence Riemann integrable.
Is the following true? $$\int_c^dg(t)f(t)dt = [g(d)F(d)-g(c)F(c)]- \int_c^d g'(t)F(t)dt\\for\ \ d,c\in[a,b]$$ i.e. can I use some "sort" of integration by parts? Thanks in advance.

p.s. I have no exposure to measure theory.

Focus
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3 Answers3

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Strong sufficient conditions for integration-by-parts found in calculus books are $F$ and $g$ have continuous derivatives. If $f$ is continuous we can be sure that $F' = f$ and it follows that for Riemann integrals

$$\tag{*}\int_c^dg(t) f(t) \, dt = g(d)F(d) - g(c)F(c) - \int_c^d F(t) g'(t) \, dt.$$

Weaker conditions for Lebesgue integrals are that $F$ and $g$ are absolutely continuous in which case (*) is true for Lebesgue integrals.

Assuming you want to work with Riemann integrals, the result is also true with your conditions. We have to be careful because the assumption is only that $f$ is Riemann integrable.

Here is a way to prove this using Riemann-Stieltjes integration and the more general integration-by-parts result for such integrals. Since $F$ has bounded variation and $g$ is continuous we have that $g$ is R-S integrable with respect to $F$ and

$$\tag{**}\int_c^d g(t) f(t) \, dt = \int_c^dg dF = g(d)F(d) - g(c) F(c) - \int_c^d F \, dg.$$

Since $g$ is continuously differentiable your result follows. Using the mean value theorem of differential calculus it can be shown that

$$\int_c^dF\, dg = \int_c^d F(t) g'(t) \, dt.$$

A bit more work is needed to justify all of the steps. The first equality in (**) can be proved using Riemann-Stieltjes sums. The second equality is integration-by-parts for R-S integrals.

RRL
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  • How can we get F is of bounded variation? Can u give some elaboration on the steps used? Im not very familiar with integrators- shouldn't the integrator be an increasing fct? – Focus Oct 05 '17 at 05:15
  • An increasing integrator is of course one way R-S integration is developed. The more general form is for integrators of bounded variation because such functions are always the difference of two increasing functions. So it's easy to prove that $F$ is of bounded variation since $F(x) = \int_a^x f^+(t) , dt - \int_a^x f^-(t) , dt$ using the positive and negative parts of $f$. – RRL Oct 05 '17 at 05:22
  • Proving integration by parts without the strongest conditions of basic calculus requires some effort one way or another. This is the gentlest way I know since we don't rely heavily on Lebesgue integrability. Everything above can be proved using limits of Riemann-Stieltjes sums. – RRL Oct 05 '17 at 05:24
  • Very nice use of Riemann-Stieltjes integral. Your approach is very easy to follow.+1 – Paramanand Singh Oct 05 '17 at 05:29
  • @RRL I don't know if I'm understanding correctly, but is $f^{-}$ defined as $|f|$ for $f<0$ and $0$ for $f>0$? (the same Q goes for $f^{+}$) If then, is $f^{+}$ or $f^{-}$ still integrable? If it is the case, I see $F$ is difference of two increasing functions. Is this condition enough for $F$ to be of bounded variation? – Focus Oct 05 '17 at 06:18
  • Yes $f^+ = \max(f,0)$ and $f^- = \max(-f,0)$. Both are nonnegative and integrable since the max of two integrable functions is integrable. The definite integrals of these parts over$[a,x]$ must then be nondecreasing with respect to $x$. Just using the definition of BV we know that a monotone function is of BV and the difference of two BV functions is of BV. If $F = G-H$ then $\sum_{j=1}^n |F(x_j) - F(x_{j-1})| \leqslant \sum_{j=1}^n |G(x_j) - G(x_{j-1})| + \sum_{j=1}^n |H(x_j) - H(x_{j-1})| .$ – RRL Oct 05 '17 at 06:25
  • Wow, thanks. I get the details now! – Focus Oct 05 '17 at 06:45
  • I feel bad nagging you but I have one final question! Regarding first equality of (**), it is not clear to me how RS sum for $\int_c^d gdF$ gets close to RS sum for $\int_c^d g(t)f(t)dt$ by taking an appropriate partition of $[c,d]$ . – Focus Oct 05 '17 at 10:53
  • Note $S(P,g,F) = \sum_{j=1}^n g(\xi_j)(F(x_j)-F(x_{j-1}) = \sum_{j=1}^n g(\xi_j)\int_{x_{j-1}}^{x_j} f(t) , dt$ and $|S(P,g,F) - \int fg| = \left|\sum_{j=1}^n \int_{x_{j-1}}^{x_j} (g(\xi_j) - g(t))f(t) , dt\right| \leqslant \sum_{j=1}^n \int_{x_{j-1}}^{x_j} |g(\xi_j) - g(t)||f(t)| , dt$. Now show the difference can be made small using boundedness of $f$ and integrability of $g$. – RRL Oct 05 '17 at 18:16
  • $\sum_{j=1}^n \int_{x_{j-1}}^{x_j} |g(\xi_j) - g(t)||f(t)| ,dt \leq M \sum_{j=1}^n \int_{x_{j-1}}^{x_j} |g(\xi_j) - g(t)| ,dt$. From here, I see continuity of $g$ will solve the problem easily, but can we conclude just from integrability of $g$ that this can be made arbitrary small? What I only get from integrability of $g$ is $|\sum_{j=1}^n \int_{x_{j-1}}^{x_j} (g(\xi_j) - g(t)) ,dt|$ is arbitrarily small for any tagged partition with mesh sufficiently small. – Focus Oct 06 '17 at 02:10
  • For $t \in I_j = [x_{j-1},x_j]$ we have $|g(\xi_j)-g(t)| \leqslant \sup_{x \in I_j}g(x) - \inf_{x \in I_j} g(x) = M_j - m_j$. Hence $\int_{x_{j-1}}^{x_j}|g(\xi_j) - g(t)| , dt \leqslant (M_j - m_j)(x_j - x_{j-1})$ and $|S(P,g,F) - \int fg | \leqslant M(U(P,g) - L(P,g))$. If $g$ is merely integrable, there is a partition $P_\epsilon$ so that for any refinement $P$ we have $U(P,g) - L(P,g) < \epsilon/M$. Thus $\int g, dF = \int fg$. – RRL Oct 06 '17 at 02:45
  • Yup. Now, I got it all figured it out. THANK YOU for your time and effort. – Focus Oct 06 '17 at 03:21
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This is not really an answer as it uses Lebesgue integration.

The result is true, but takes work to prove using Riemann apparatus.

However, it is straightforward to see using the Lebesgue integral:

Since $f$ is Riemann integrable it is bounded and hence $F$ is absolutely continuous (ac.) and $F'(x) = f(x) $ for ae. $x$. Since $g$ is $C^1$ it is ac. and hence the function $\phi(x) = g(x) F(x)$ is ac. We see that $\phi'(x) = g'(x) F(x) + g(x) f(x)$ for ae. $x$.

Hence we have the desired result (except that we are using the Lebesgue integral): $\phi(d)-\phi(c) = \int_c^d \phi'(x) dx = \int_c^d g'(x) F(x)dx + \int_c^d g(x) f(x) dx$.

To finish, note that the functions $x \mapsto g'(x) F(x)$, $x \mapsto g(x) f(x)$ are bounded and continuous ae. hence the integrals are the same as the Riemann integral.

copper.hat
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Try integration by parts:

$$\int_c^d g(t)f(t)dt $$ $$=\int_a^d g(t)f(t)dt - \int_a^c g(t)f(t)dt$$ $$=\left.\left[g(t)\int f(t)dt - \int g'(t)\left\{\int f(t) dt\right\} \right]\right|_a^d - \left.\left[g(t)\int f(t)dt - \int g'(t)\left\{\int f(t) dt\right\} \right]\right|_a^c$$ $$\ldots \ldots$$ $$\ldots \ldots$$ $$= [g(d)F(d)-g(c)F(c)]- \int_c^d g'(t)F(t)dt$$

Try to fill up the gaps. Its easy.

Hope this helps you.

Alex S
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SchrodingersCat
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  • But we dont know if F is differentiable so the "ordinary" integration by parts fails – Focus Oct 05 '17 at 04:57
  • How do you know that $F' = f$ when all that is assumed is that $f$ is Riemann integrable not continuous? Under what conditions is integration by parts as you are writing it true. – RRL Oct 05 '17 at 04:57
  • @HeeJuneKim I dont see my solution involving any differentiation of F. I dont know where you saw it. – SchrodingersCat Oct 05 '17 at 05:00
  • Your solution assumes that $F$ (given by $\int f(t) , dt$) is anti-derivarive of $f$. And this is not guaranteed for all Riemann integrable functions $f$. – Paramanand Singh Oct 05 '17 at 05:32
  • Ok I will delete this... but now I am on app. I will do it later. – SchrodingersCat Oct 05 '17 at 06:11