Find a function f so that:

$\int _a^{x^2}\:f\left(t\right)ln\left(t\right)dt\:=\:x^3\left(ln\left(x\right)-\frac{1}{3}\right)$, $a>1$

This is how I did it:

First get the derivative of $x^3\left(ln\left(x\right)-\frac{1}{3}\right)$ which is $3x^2ln(x)$

Then the derivative of the integral which according to the theorem is: $2xf(x^2)ln(x^2) - f(a)ln(a)*a'$, but $a'=0$

Now solve for $f(x)$, $2xf(x^2)ln(x^2) = 3x^2ln(x)$

$f(x^2) =\frac{3x^2ln\left(x\right)}{2xln\left(x^2\right)}$

$f(x^2) = 3x/4$

That makes $f(x) = \frac{3\sqrt{x}}{4}$, right?

  • $\begingroup$ I got $f(x) = \frac{3}{2}x$, but I didn't plug it back in to check if it works. $\endgroup$ – Logan Clark Nov 3 '16 at 0:51
  • $\begingroup$ @LoganClark are you 100% sure? $\endgroup$ – bajotupie Nov 3 '16 at 1:11
  • $\begingroup$ It can't be true that $f(x)$ is independent of $a$. Take $a=4$ and $x=2$ in the original expression. Then since $\int_a^{x^2}f(t)\log(t)dt=\int_4^4 3/4\sqrt{t} \log(t) dt = 0$ and $x^3(\log(x)-1/3)=8(\log(2)-1/3)\neq 0$ it appears your answer is not correct? What appears to be the mistake is that if $f(t)\log(t)=\mathrm{constant}$ or $f(t)=g(a)/\log(t)$ then the constant vanishes. So you can add $g(a)/log(t)$ to the function $f$ without disturbing your derivative equation. We need to find $g(a)$ to match the LHS and RHS for a value of $x$ for which the integral can be evaluated. $\endgroup$ – Peder Nov 3 '16 at 2:05
  • $\begingroup$ Im confused now hahah $\endgroup$ – bajotupie Nov 3 '16 at 2:08
  • $\begingroup$ Sorry for adding a deleting comments at a maddening pace. I think it is true that your answer is incorrect, but my argument in the comment above is not exactly correct. Mathematica says that the indefinite integral of $\sqrt{t}\log(t)$ is $2/9 t^{3/2} (-2+3 \log(t))$, so we can verify that your answer is incorrect by direct computation. The problem seems to be that when you take the derivative of the LHS and RHS constants vanishes. You have to figure out how to put these constants back in to get the right answer. $\endgroup$ – Peder Nov 3 '16 at 2:42

Note that if you take $f(t) = \frac{g(a)}{(x^2-a)\log(t)}$ then $$ \int_{a}^{x^2} f(t)\log(t)dt = \int_{a}^{x^2} \frac{g(a)}{(x^2-a)}dt = g(a) $$ Moreover, if you take the derivative of $g(a)$ with respect to $x$ you will get $0$, so that means that when you took derivatives of the left hand side and the right hand side you have to take into a account that you can add any term $\frac{g(a)}{(x^2-a)\log(t)}$ to the function $f(t)$ without the derivative with respect to $x$ of the LHS changing.

That being said you did all the hard work. Since Mathematica gives the integral $$ \int_0^x \sqrt{t}\log(t) dt = \frac{2}{9}x^{3/2}(3\log(x)-2) $$ we can plug in $x^2$ into this expression to verify that $$ \int_0^{x^2} \frac{3}{4}\sqrt{t}\log(t) dt =x^3(\log(x)-\frac{1}{3}),$$ so your answer is the solution when $a=0$.

We can find the general solution by noting that $$ \int_a^{x^2} \left(\frac{3}{4}\sqrt{t}+\frac{g(a)}{(x^2-a)\log(t)}\right) \log(t) dt = g(a)+x^3(\log(x)-\frac{1}{3})-a^{3/2}(\frac{1}{2}\log(a)-\frac{1}{3}) $$ The answer should be $x^3(\log(x)-\frac{1}{3})$ therefore we must have $g(a) = a^{3/2}(\frac{1}{2}\log(a)-\frac{1}{3})$ which means that $$ f(t) = \frac{3}{4}\sqrt{t}+\frac{a^{3/2}(3\log(a)-2)}{6(x^2-a)\log(t)} $$ You had the requirement $a>1$ in your problem statement. It seems the statement holds for any $a\geq 0$ by my argument.

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