Interesting Integral including $\ln x$

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:-)

Solution 1:

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}

Solution 2:

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.