# Evaluating this definite integral

Given: $$I=\int_0^1 \frac{\sin(\ln(x))}{\ln(x)}dx$$

Now I have tried every strategy, I could think of, substitution, by parts (it will return the same $$I$$ after a few times), etc.

In the solution provided, they did introduce fancy complex numbers, which I have no idea of. Note that I am a normal high school student, so I cannot understand high-level calculus. Please help here.

• Wolfy might give you a clue ? wolframalpha.com/… Nov 16, 2022 at 13:26
• @DonaldSplutterwit, I checked but they also introduced those complex numbers. Is there no other way to do this by traditional methods (which I need). Nov 16, 2022 at 13:29
• Sub $u=- \log(x)$ ... Taylor series for $\sin$ ... integrate term by term ... Alternating sum of harmonic odd numbers = ? Nov 16, 2022 at 13:35
• Use differentiation under the integral sign. Nov 16, 2022 at 13:44
• @DonaldSplutterwit's suggested substitution can be exploited another way, by noticing a Frullani integral viz. $\frac{1}{2i}\ln\frac{1+i}{1-i}=\frac{\pi}{4}$.
– J.G.
Nov 16, 2022 at 16:49

Consider the integral $$I(\alpha) = \int\limits_{0}^{1} \frac{\sin{\left(\alpha \log{\left(x \right)} \right)}}{\log{\left(x \right)}}\, dx$$ Your integral becomes $$I(1)$$. Note that $$I(0) = \int\limits_{0}^{1} \sin(0)\, dx = 0 \tag{\dagger} \label{eq}$$ Let's differentiate $$I(\alpha)$$ with respect to $$\alpha$$
$$I'(\alpha) = \int\limits_{0}^{1} \cos{\left(\alpha \log{\left(x \right)} \right)}\, dx\\ = \left.\frac{\alpha x \sin{\left(\alpha \log{\left(x \right)} \right)}}{\alpha^{2} + 1} + \frac{x \cos{\left(\alpha \log{\left(x \right)} \right)}}{\alpha^{2} + 1}\right|_{x=0}^{x=1} = \frac{1}{\alpha^{2} + 1}$$ So integrating back we get: $$I(\alpha) = \arctan(\alpha) + C$$ But remember we know by $$\ref{eq}$$ that $$I(0) = 0$$ so $$0 = \arctan(0) + C = 0 + C \implies C = 0$$ So $$I(\alpha) = \arctan(\alpha)$$ and specifically, your integral becomes $$I(1) = \arctan(1) = \frac{\pi}{4}$$
Using $$\sin x=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}, \int_0^1(\ln x)^ndx=(-1)^nn!$$ one has $$\begin{eqnarray} I&=&\int_0^1 \frac{\sin(\ln(x))}{\ln(x)}dx\\ &=&\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n-1}(\ln x)^{2n-2}}{(2n-1)!}dx\\ &=&\sum_{n=1}^\infty\frac{(-1)^{n-1}(2n-2)!}{(2n-1)!}\\ &=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}\\\\ &=&\frac\pi4. \end{eqnarray}$$