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.

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

2 Answers 2


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 $$I(1) = \arctan(1) = \frac{\pi}{4}$$

The lesson: When in doubt, remember Feynman.

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    $\begingroup$ Thank you very much. I almost forgot of reduction strategy and subsequent method is pretty new to me. Thank you again $\endgroup$ Nov 16, 2022 at 13:46

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}


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