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I would like to evaluate this integral: $$\int_0^1 \frac{\sin(\ln(x))}{\ln(x)}\,dx$$

I tried a lot, started by integral uv formula [integration-by-parts?] and it went quite lengthy and never ending.

Then, thought may be expansion of $\sin x$ but it didn't made any sense, it's a infinite series and there's no way here that it will end..

It is highly appreciated if you give some hints to solve it.

I am a high school student,so I expect it to be simple but tricky:-)

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  • $\begingroup$ Have you tried substitution? $\endgroup$ Apr 20 '21 at 17:44
  • $\begingroup$ Yes, I tried that also, but what can be Substituted? $\endgroup$
    – Rover
    Apr 20 '21 at 17:46
  • $\begingroup$ There's not a nice solution to this, the antiderivative requires exponential integrals. Now, if it were $\int \frac{\sin(\ln(x))}{x}\,dx$ that would be a different matter. $\endgroup$
    – DMcMor
    Apr 20 '21 at 17:47
  • $\begingroup$ @DMcMor : Okay , if limit are not there to integral , how can it be solved? After integrating can't we apply limits ? $\endgroup$
    – Rover
    Apr 20 '21 at 17:50
  • $\begingroup$ @Rover Series expansion works, and is possibly the most elementary one if we calculate $\int_0^1 \ln^n x \, \mathrm dx$ via integration by parts. See the third solution in my answer. $\endgroup$
    – Zack
    Apr 20 '21 at 19:56
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Observe that $\displaystyle \frac{\sin(\ln(x)) }{\ln{x}} = \int_0^1 \cos(t \ln{x}) \, \mathrm dt$. Then:

$$\begin{aligned} \int_0^1\frac{\sin(\ln(x)) }{\ln{x}}\, \mathrm dx &= \int_0^1 \int_0^1 \cos(t\ln{x})\;{\mathrm dt}\;{\mathrm dx} \\& = \ \int_0^1 \int_0^1 \cos(t\ln{x})\;{\mathrm dx}\;{\mathrm dt} \\&= \int_0^1 \frac{1}{t^2+1} \;{\mathrm dt} \\& = \frac{\pi}{4}. \end{aligned}$$


Equivalently, consider the function

$$\displaystyle f(t) = \int_0^1\frac{\sin(t\ln(x)) }{\ln{x}}\, \mathrm dx.$$

Then

$$\displaystyle f'(t) = \int_0^1 \cos(t \ln{x})\, \mathrm{d}x = \frac{1}{1+t^2}.$$

Therefore $f(t) = \arctan(t)+C$. But $f(0) = 0$ so $C = 0$.

Hence $f(t) = \arctan(t)$. We seek $\displaystyle f(1) = \arctan(1) = \frac{\pi}{4}$.


Series solution:

$\displaystyle I = \int_0^1\frac{\sin(\ln(x)) }{\ln{x}}\, \mathrm dx = \int_0^1\sum_{k \ge 0} \frac{(-1)^k \ln^{2k}{x}}{(2k+1)!}\, \mathrm dx = \sum_{k \ge 0} \frac{(-1)^k }{(2k+1)!} \int_0^1 \ln^{2k}{x}\, \mathrm dx $

Then we calculate $\displaystyle \int_0^1 \ln^nx \,\mathrm{d}x$ via integration by parts to find that it's equal to $(-1)^n n!$

Or consider $\displaystyle f(m) = \int_0^1 x^m \,{dx} = \frac{1}{1+m}.$

Then taking the $n$-th derivative of both sides:

$\displaystyle f^{(n)}(m) = \int_0^1 x^m \ln^{n}{x} \,{dx} = \frac{(-1)^n n! }{(1+m)^{n+1}}.$

In either case we get $\displaystyle \int_0^1 \ln^{2k}{x}\, \mathrm dx = (2k)!$. Hence:

$\displaystyle I = \sum_{k \ge 0} \frac{(-1)^k (2k)!}{(2k+1)!} = \sum_{k \ge 0} \frac{(-1)^k (2k)!}{(2k+1)(2k)!} = \sum_{k \ge 0} \frac{(-1)^k }{(2k+1)} = \frac{\pi}{4}.$

To prove the last equality, consider $\displaystyle \frac{1}{2k+1} = \int_0^1 x^{2k} \, \mathrm dx$

and the geometric series $\displaystyle \sum_{k \ge 0}(-1)^kx^{2k} = \frac{1}{1+x^2}$. Then

$\begin{aligned} \displaystyle \sum_{k \ge 0} \frac{(-1)^k}{2k+1} & = \sum_{k \ge 0}{(-1)^k}\int_0^1 x^{2k}\,{\mathrm dx} \\& = \int_0^1 \sum_{k \ge 0}(-1)^k x^{2k} \, \mathrm dx \\& = \int_0^1 \frac{1}{1+x^2}\,\mathrm dx \\& = \frac{\pi}{4}.\end{aligned}$


Regarding the integral $\displaystyle I = \int_0^1 \cos(t \ln{x})\, \mathrm{d}x $, we let $x = e^{-y}$. Then $\displaystyle I = \int_0^\infty e^{-y}\cos(ty)\,\mathrm{d}y.$ We get the answer by applying integration by parts (twice). Or we can consider the real part, if we're familiar with complex numbers:

\begin{align} I & = \int_0^\infty e^{-y}\cos(ty)\,\mathrm{d}y =\Re\left(\int_0^\infty e^{-(1-it)y}\mathrm{d}y\right)\\ &=\Re\left(\int_0^\infty e^{-(1+t^2)y}\mathrm{d}(1+it)y\right)\\ &=\Re\left(\frac{1+it}{1+t^2}\int_0^\infty e^{-(1+t^2)y}\mathrm{d}(1+t^2)y\right)\\ &=\Re\left(\frac{1+it}{1+t^2}\right)\\& =\frac{1}{1+t^2}. \end{align}

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    $\begingroup$ Nice argument!!! $\endgroup$
    – Igor Rivin
    Apr 20 '21 at 18:22
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    $\begingroup$ (+1) both nice approaches with functions for OPs stated level, nicely done! $\endgroup$
    – Henry Lee
    Apr 20 '21 at 18:52
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    $\begingroup$ @Rover That's the Liebniz formula for $\pi$. Read the bit I added or see the proofs in that Wiki page. $\endgroup$
    – Zack
    Apr 21 '21 at 8:03
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    $\begingroup$ I would suggest to add a step explaining $\int_0^1\cos(t\ln x)dx=\frac1{1+t^2}$. $\endgroup$
    – user
    Apr 21 '21 at 9:06
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    $\begingroup$ Hi, in the first method, how did you solve $\int_0^1\cos(t\ln x)dx$? $\endgroup$
    – aarbee
    May 8 '21 at 11:50
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As you said, you're in high-school. So, the most obvious method is to use the substitution method and the make an observation that the resulting integral involves the Exponential Integral of complex argument.Let us start by making the substitution $-u=\ln(x).$ Then,

$\begin{equation}\begin{aligned}\int^{1}_{0}\dfrac{\sin(\ln(x)}{\ln(x)}&=\int^{\infty}_{0}\dfrac{\sin(u)e^{-u}}{u}du\\ &=\dfrac{1}{2}\int_{0}^{\infty}\dfrac{e^{-u}}{u}\left(ie^{-iu}-ie^{iu}\right)du\\ &=\dfrac{i}{2}\int^{\infty}_{0}\left(\dfrac{e^{-(i+1)u}}{u}-\dfrac{e^{-(1-i)u}}{u}\right)du\\ &=\dfrac{1}{2}i\left\{\mathbb{E}_1[-(1+i)u]-\mathbb{E}_1[-(1-i)u]\right\}|_{0}^{\infty}\\ &=\dfrac{1}{2}i\left\{\mathbb{E}_1[(1+i)\ln(x)]-\mathbb{E}_1[(1-i)\ln(x)]\right\}|_{0}^{1}\\ &=\dfrac{\pi}{4} \end{aligned} \end{equation}$

We have expanded the sine function into its complex exponential form and used the fact that $\int\dfrac{e^{-xt}}{t}dt=\mathbb{E}_1(-x),$ with $\mathbb{E}_1(\cdot)$ as the well-known Exponential Integral function of complex argument Exponential Integral.

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The integral equals $\pi/4.$ The indefinite integral does involve expontential integrals.

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    $\begingroup$ Any other method without involving exponential integrals? $\endgroup$
    – Rover
    Apr 20 '21 at 18:16
  • $\begingroup$ I recommend providing more context than just the value of the integral, as this provides little value to the OP's understanding of how to perform the calculation themselves. $\endgroup$
    – ndhanson3
    Apr 20 '21 at 22:35
  • $\begingroup$ @ndhanson3 Noted. However, it was not clear what level the OP was at. For some people, special values of exponential integral are not a problem, for some they are :) $\endgroup$
    – Igor Rivin
    Apr 21 '21 at 2:40

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