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Let $f(x)$ be smooth function on $\mathbb{R}$ and

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

I need to find $\ f(x)$.

My first thoughts were to solve this by differentiating under the integral sign, but this topic was almost not covered at the university, I know about this method only from videos on YouTube, so I'm not quite sure how to approach this problem correctly.

Edit 1: I tried to apply $\textbf{Leibniz integral rule}$ and that't what I've got. $$\frac{\partial{\int_{-x}^{x}t^2f(t)dt}}{\partial{x}}=\frac{\partial{(f(x)-1)}}{\partial{x}}$$ $$x^2f(x)-x^2f(-x)=f'(x)$$ Now I supposed that f(x) is an even function, if it wasn't true, then $\int_{-x}^{x}t^2f(t)dt$ simply was equal to $0$ So $$x^2f(x)-x^2f(x)=f'(x)$$ $$0=f'(x)$$ $$f(x) = C$$

Then I substituted that in the integral $$\int_{-x}^{x}t^2Сdt=С-1$$ $$\frac{2Cx^2}{3}=С-1$$ $$C=\frac{1}{1-\frac{2}{3}x^3}=f(x)$$ And this is false, because as we understood earlier $f(x)$ is a constant.

So where do I went wrong?

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  • $\begingroup$ Is nothing known about $\ f(t)$? Something must be given to proceed. $\endgroup$
    – user1173615
    Dec 9, 2023 at 15:40
  • $\begingroup$ I edited the question, now there is only $f(x)$ $\endgroup$
    – Ilya
    Dec 9, 2023 at 15:42
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    $\begingroup$ Apply the Leibnitz Integral Rule. en.wikipedia.org/wiki/Leibniz_integral_rule Just to confirm, is nothing else given about the nature of f(x)? $\endgroup$
    – user1173615
    Dec 9, 2023 at 15:45
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    $\begingroup$ Yes, only its smoothness. $\endgroup$
    – Ilya
    Dec 9, 2023 at 15:47
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    $\begingroup$ Basically you prooved that $f$ is not even. $\endgroup$ Dec 9, 2023 at 16:30

3 Answers 3

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Writing $f(x) = 1 + g(x)$, we get

$$ g(x) = \frac{2x^3}{3} + \int_{-x}^{x} t^2 g(t) \, \mathrm{d}t. $$

This shows that $g$ is and odd function, which then forces $\int_{-x}^{x} t^2 g(t) \, \mathrm{d}t = 0$. Therefore $g(x) = \frac{x^3}{3}$ and hence

$$ f(x) = 1 + \frac{2x^3}{3}. $$

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Replace $\ x$ with $\ -x$ throughout in the given equation, we get:

$$\ f(-x) - 1 = -(f(x)-1) = 1 - f(x)$$

Or, more importantly: $$\ f(-x) = 2 - f(x)$$

Thus, we observe that while $\ f(x)$ doesn't exactly change signs when $\ x$ is replaced by $\ -x$, $\ (f(x) - 1)$ does change its sign.

Note: I must correct a slight mistake of yours. You have claimed that

$$ x^2f(x) - x^2f(-x) = f'(x)$$

I ended up spending a lot of time figuring out why the function obtained by assuming this relation to be correct did not satisfy the original equation.

But, the correct result one obtains after applying the Leibniz Integral rule properly, would be:

$$x^2 \cdot f(x) \cdot (1) -x^2 \cdot f(-x) \cdot (-1)=f'(x)$$

It seems you forgot to multiply (-1) on the second derivative, since $\frac{d(-x)}{dx}$ is -1, not 1.

Rewriting, we get:

$$\ x^2 \cdot (f(x) + f(-x)) = \frac{d}{dx}(f(x)) $$

Now, given that $\ f(-x) = 2 - f(x)$, i.e. $\ f(x) +f(-x) =2$, we rewrite:

$$\frac{d}{dx}(f(x) = 2x^2$$

And now we finally obtain

$$f(x) = \frac{2x^3}{3} + C$$

We obtain C by observing that at $\ x = 0$, $\ f(x)$ is equal to 1.

Thus, $$\ f(x) = 1 + \frac{2x^3}{3} $$ as observed earlier.

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    $\begingroup$ Thanks, you're great! $\endgroup$
    – Ilya
    Dec 9, 2023 at 17:14
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You can replace x ← -x and get

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

$\iff -f(x)+1=f(-x)-1$

$\iff f(x)+f(-x)=2$

$\iff x^2f(x)+x^2f(-x)=2x^2$

$\iff \int_{-x}^{x}t^2f(t)+t^2f(-t)=\frac{4x^3}{3}$ Call this relation 1

Now take the integral

$\int_{-x}^{x}t^2f(t)dt$

do the substitution t=-y and we get

$\int_{-x}^{x}y^2f(-y)dy=\int_{-x}^{x}t^2f(t)dt$

$\iff \int_{-x}^{x}y^2(f(y)-f(-y))dy=0$ If we add relation 1 to this we get

$\iff 2\int_{-x}^{x}y^2f(y)dy=\frac{4x^3}{3}$

$\iff 2(f(x)-1)=\frac{4x^3}{3}$

$\iff f(x)=\frac{2x^3}{3}+1$ which indeed satisfies the initial condition

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