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Considering this trivial definite integrals inequality: $$ \mbox{with} \;a \in (0, 2\pi)\; \mbox{such that} \;\; \int_{0} ^{a} \cos(x) dx \; = \; 0 \; \; \; \;\Longrightarrow \; \; \; \; \int_{0} ^{a} \sin(x) dx \; \neq \; 0 $$ Inserting now a multiplying function $f(x)$ (a not identically zero continuous real valued function of real variable), and after having tried to visualize the evolution of the total signed area for various $f(x)$ examples, it would appear to me that also the following inequality might hold: $$ \mbox{with} \;a\; \mbox{such that} \;\; \int_{0} ^{a} \cos(x)\;f(x) \; dx \; \; = \; 0 \; \; \;\Longrightarrow \; \; \; \; \int_{0} ^{a} \sin(x) \;f(x) \; dx \; \neq \; 0 $$ is this true ?

NOTE: the last edit is posted as a new question, because I have accepted @5xum answer and said last edit has now changed the question to an extent which would require a different answer.

Luca
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  • There are many functions satisfying $\int_0^{2\pi} \cos(x)f(x)dx = \int_0^{2\pi} \sin(x)f(x)dx = 0$ – think of Fourier coefficients! – Martin R May 17 '21 at 08:14
  • What do you mean by your "$\mbox{with} ;a; \mbox{such that} ;; \int_{0} ^{a} \cos(x) dx ; = ; 0 ; ; ; ;\Longrightarrow ; ; ; ; \int_{0} ^{a} \sin(x) dx ; \neq ; 0$?" Do you mean $a$ is a fixed number which satisfy both conditions? – Arctic Char May 17 '21 at 08:40
  • @Arctic Char indeed you are right, I shall have excluded $a=0$ and $a=2k\pi$. See whether reformulating it to restrict $a$ to $(0, 2\pi)$ has now fixed the error. – Luca May 17 '21 at 09:20

1 Answers1

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It is not true, for example, you could have $\forall x: f(x)=0$, in which case both integrals would be $0$.

For a less trivial example, take any function $g(x)$, and define $$f(x) = g(x) - A\cos(x) - B\sin(x)$$ where $$A=\frac{1}{\pi} \int_0^{2\pi} \cos(x) g(x) dx$$ and $$B= \frac{1}{\pi} \int_0^{2\pi} \sin(x)g(x)dx.$$

Then, you have

$$\begin{align}\int_0^{2\pi} f(x)\cos(x)dx &=\int_0^{2\pi} g(x)\cos(x)dx - A\int_{0}^{2\pi}\cos^2(x)dx - B\int_{0}^{2\pi}\cos(x)\sin(x)dx \\&=\int_0^{2\pi} g(x)\cos(x)dx - A\cdot \pi - B\cdot 0\\ &=\int_0^{2\pi} g(x)\cos(x)dx - \pi\cdot \frac{1}{\pi} \int_0^{2\pi} \cos(x) g(x) dx = 0\end{align}$$

and you would also have

$$\begin{align}\int_0^{2\pi} f(x)\sin(x)dx &=\int_0^{2\pi} g(x)\sin(x)dx - A\int_{0}^{2\pi}\sin(x)\cos(x)dx - B\int_{0}^{2\pi}\sin^2(x)dx \\&=\int_0^{2\pi} g(x)\sin(x)dx - A\cdot 0 - B\cdot \pi\\ &=\int_0^{2\pi} g(x)\sin(x)dx - \pi\cdot \frac{1}{\pi} \int_0^{2\pi} \sin(x) g(x) dx = 0\end{align}$$

and this presents a counterexample to your question.

5xum
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  • well, why would anybody be interested in the case integral of $0$ = integral of $0$ ? – Luca May 17 '21 at 08:13
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    @Luca I don't know, but you asked if $\int_0^a\cos (x)f(x)dx=0\implies \int_0^a\sin(x)f(x)dx$ is a true stattement, and I provided an example when this is not a true statement. I don't know if it's an "interesting" counterexample, but it is a counterexample, is it not? – 5xum May 17 '21 at 08:14
  • But $a = 2\pi$ does not work as it does not satisfy the assumption. – Arctic Char May 17 '21 at 08:29
  • @5xum I might accept it as a comment about the general approach that one shall always try to be exceedingly rigorous, but not as an answer. The "accept as an answer button" says "accept this answer if it solved your problem or was the most helpful in finding your solution", but yours does neither. – Luca May 17 '21 at 08:31
  • @Luca I answered the question you posed exactly to highlight how your question was probably not the one you wanted to ask. Anyway, I expanded my answer, and it's still "no". – 5xum May 17 '21 at 08:32
  • @ArcticChar Which assumption? – 5xum May 17 '21 at 08:32
  • $\int_0^a \cos x dx = 0$ and $\int_0^a \sin x dx \neq 0$. – Arctic Char May 17 '21 at 08:34
  • @ArcticChar I don't see that this would be an assumption. I don't read the question the same way, I guess? – 5xum May 17 '21 at 08:35
  • @5xum OK, thanks. I have finally accepted it as an answer, as it helped me in identifying a more serious problem in my "trivial example" opening: I shall have excluded $a=0$ and $a=2 \pi$. So, your answer is accepted, and I have now updated my original question. I am not sure whether the rules would require me to instead post a new question. – Luca May 17 '21 at 09:05
  • @Luca It's better to ask a new question in that case. The questions on the site are meant to be fairly self contained, and they should not be lengthy discussions with several subquestions after the first one. – 5xum May 17 '21 at 09:17