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?

Considering the integral
$$
I(a)=\int_0^1 \frac{\sin (a \ln x)}{\ln x} d x
$$
Letting $y=-\ln x$ and then differentiating $I(a)$ w.r.t. $a$ gives
$$
\begin{aligned}
I^{\prime}(a) & =\int_{\infty}^0 \cos (-a y)\left(-e^{-y} d y\right. \\
& =\int_0^{\infty} e^{-y} \cos (a y) d y \\
& =\frac{1}{a^2+1}
\end{aligned}
$$
Integrating $I^{\prime}(a)$ from $0$ to $1$ gives
$$
I(1)-I(0)=\int_0^1 \frac{1}{u^2+1} d u=\frac{\pi}{4}
$$
Hence $$\boxed{\int_0^1 \frac{\sin (\ln x)}{\ln x} d x=I(1)= \frac{\pi}{4}}$$
