Proof that $\int_0^1\frac{\sin\ln x}{\ln x}dx=0$ is wrong So I proved that $$\int_0^1\frac{\sin\ln x}{\ln x}dx=0$$Which is wrong according to Desmos. What is not right about my proof?

Let $$I(a)=\int_0^1\frac{\sin\ln x}{a\ln x}dx$$Then$$I'(a)=-\int_0^1\frac{\sin\ln x}{a^2\ln x}dx\implies-aI'(a)=I(a)$$Integrate both sides: $$-\int aI'(a)da=^{\text{integration by parts}}-aI(a)+\int I(a)da=\int I(a)da\implies-aI(a)=0$$

I looked over it and couldn't find anything wrong. Maybe I messed up in the integration by parts?

Since you took an indefinite integral of both sides, you should get
$$ -a I(a) + \int I(a)\, da = \int I(a)\, da + C $$
$$ a I(a) = -C $$
and this just shows $a I(a)$ is a constant number, but that was obvious from the definition of $I(a)$.
(Trying a definite integral has just the same issue: For all $a_1, a_2 \in \mathbb{R}$,)
$$ -a_2 I(a_2) + a_1 I(a_1) + \int_{a_1}^{a_2} I(a)\, da = \int_{a_1}^{a_2} I(a)\, da $$
$$ a_1 I(a_1) = a_2 I(a_2) $$
