My brother brought me this same question, also on this website.

$$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt \tag{1}$$

When I try to solve it, I also did same thing. Differentiation leads to

$$f(x) = 1-xf(x)\\ f(x) = \frac{1}{1+x} \\ \int_{0}^{1}f(x) dx = \ln 2 $$

But try putting $x = 1$ in original equation $(1)$,

$$\int_{0}^{1} f(t) dt = 1$$

so what has happened? We get two contradictory results. Is it still true that $f(1) = 1/2$? Where is the mistake?

  • $\begingroup$ This reminds me of an old question from India's prestigious JEE for admission to top government engineering colleges. See math.stackexchange.com/q/1495336/72031 $\endgroup$ – Paramanand Singh Mar 12 '18 at 6:31
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    $\begingroup$ +1 to you for checking that $f(x) =1/(1+x)$ leads to a contradiction. Most students stop at the answer $f(x) =1/(1+x)$. $\endgroup$ – Paramanand Singh Mar 12 '18 at 6:34
  • $\begingroup$ Wow again we have 3 close votes. Where are my popcorns? $\endgroup$ – King Tut Mar 15 '18 at 16:23

You are correct. The problem is with the person who made the question.

The functional equation is incorrect.

The correct functional equation for this function must be :

$$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt + \ln 2 -1$$

There are always issues, if you are differentiating an expression, and there are constants.

Since taking derivative destroys existence of any constant, often there are mistakes in questions, if not formatted correctly.

  • $\begingroup$ Does that imply that original functional equation has no solutions? $\endgroup$ – wilkersmon Feb 14 '18 at 0:02
  • $\begingroup$ @wilkersmon Yes. Definitely. $\endgroup$ – Jaideep Khare Feb 14 '18 at 5:09
  • $\begingroup$ @KingTut Check this out $\endgroup$ – Jaideep Khare Mar 10 '18 at 14:03
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    $\begingroup$ +1 for the starting line. Settings questions for an exam/book is not as easy as it may appear. $\endgroup$ – Paramanand Singh Mar 12 '18 at 6:38

If two functions have the same derivative, all you can say about them is that they differ by a constant. So the $f$ you found is the solution to an equality of the form $$ \int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt + c, $$ where $c$ is not necessarily $0$. In particular, by differentiating you lost the chance of identifying $f(1)$.

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    $\begingroup$ Well... $c = \ln 2 -1 $, atleast for this case. $\endgroup$ – Jaideep Khare Feb 13 '18 at 14:45
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    $\begingroup$ Indeed. $\ \ \ \ $ $\endgroup$ – Martin Argerami Feb 13 '18 at 14:46
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    $\begingroup$ I'm not sure what you are aiming for. If you differentiate, you lose constants. If you square, you lose minus signs. They are things one needs to be aware of when processing an equation. $\endgroup$ – Martin Argerami Feb 13 '18 at 14:56
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    $\begingroup$ Yes, now i understand. $\endgroup$ – King Tut Feb 13 '18 at 15:48

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