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This problem $5b$ from Chapter $14$ of Spivak's Calculus:

Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ (Note $g$ is not assumed to be continuous at $0$).

I think this question is impossible? First of all there's no restrictions on $x$, so letting $x = 1$ and $x = -1$, we get $0 = 2$.

So maybe we have to restrict to $x \geq 0$? In that case, we can take the square root:

Find $g$ such that $\forall x \geq 0 : \int_0^{x}tg(t)dt = x + \sqrt{x}$

Let $f(t) = tg(t)$, so we need $\int_0^{x}f(t)dt = x + \sqrt{x}$

Now $\frac{d}{dx} (x + \sqrt{x}) = 1 + \frac{1}{2\sqrt{x}}$. But $f(t) = 1 + \frac{1}{2\sqrt{t}}$ won't work because $1 + \frac{1}{2\sqrt{t}}$ isn't bounded as $t \rightarrow 0$, so the integral won't be defined.

Am I missing something?

VIVID
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SenZen
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2 Answers2

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The function $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3}$$ seems to satisfy the condition. Below is how to find it:

Let $G(x) = \int xg(x)dx$. Then, we have $$\int_0^{x^2}tg(t)dt \overset{\text{(FTC)}}= G(x^2) - G(0) \overset{*}= x + x^2 $$ Now, differentiate both sides of $(*)$ to get $$\begin{align} 2x[x^2g(x^2)] &= 1 + 2x \\[1mm] 2x^3g(x^2) &= 1+ 2x \\ g(x^2) &= \frac{1+2x}{2x^3}.\end{align}$$ Therefore, we get $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3} $$

Indeed, $$\int_0^{x^2} t \frac{1 + 2\sqrt t}{2(\sqrt t)^3} dt = \int_0^{x^2}\left( 1 + \frac{1}{2\sqrt t}\right)dt = \left[\sqrt t + t\right]_0^{x^2} = x + x^2$$ as desired.

VIVID
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  • The last step is wrong? $\int_0^{x^2} t \frac{1 + 2\sqrt t}{2(\sqrt t)^3} dt$ doesn't exist, because the function $f(t) = t \frac{1 + 2\sqrt t}{2(\sqrt t)^3}$ isn't bounded on $[0, x]$ for any $x$? – SenZen Dec 29 '20 at 14:31
  • @SenZen Note that the unboundedness of a function does not always imply that the integral does not exist, for example, see these examples at Wiki: https://en.wikipedia.org/wiki/Improper_integral#Examples – VIVID Dec 29 '20 at 14:57
  • Ok but at this point in the book, Spivak hasn't talked about extending the integral to unbounded functions using limits, so it can't be the answer he's looking for. Specifically if we restrict the problem to bounded functions $g$ such that $xg(x)$ is bounded on $[0, \infty)$, does there exist a solution? – SenZen Dec 29 '20 at 15:25
  • @SenZen But its primitive is bounded in $[0,x^2]$ for any $x$. – VIVID Dec 29 '20 at 15:30
  • VIVID, it seems to me that @SenZen is correct: $tg(t)$ is unbounded on every interval $(0,\varepsilon)$. – TonyK Dec 29 '20 at 15:32
  • @SenZen What has your book covered so far? That improper integral is convergent, you can see some discussions about it here: https://math.stackexchange.com/questions/3015748/is-1-sqrtx-riemann-integrable-on-0-1?noredirect=1 – VIVID Dec 29 '20 at 15:36
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    @SenZen, if $tg(t)$ is bounded on $(0,1]$, say $|tg(t)|\le M$ (and here we can choose $M>1$ for convenience), then $|\int_0^{x^2}tg(t)dt|\le Mx^2<x+x^2$ for $x<1/(M-1)$. So $tg(t)$ has to be unbounded. – TonyK Dec 29 '20 at 15:39
  • @TonyK Yes, neither do I say that it is bounded there but the improper integral converges and $g$ satisfies the problem :) – VIVID Dec 29 '20 at 15:40
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If we let: $$F(x)=\int tg(t)dt$$ then we have: $$F(x^2)-F(0)=x+x^2$$ now try differentiating both sides: $$2xF'(x^2)=1+2x$$ but we know that $F'(x)=xg(x)$ and so: $$2x\left[x^2g(x^2)\right]=1+2x$$ $$2x^3g(x^2)=1+2x$$ $$g(x^2)=\frac{1}{2x^3}+\frac{1}{x^2}$$ now find $g(x)$

Henry Lee
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    It's strange that Spivak worded the question so as to be misleading as such. He could've said "does there exist a $g$..." and then add a part (c) "does $\int_0^1 tg(t)$ exist?" I wonder what percentage of people who did Spivak's book "solved" the question this way and didn't realize the fallacy. – SenZen Dec 30 '20 at 07:54