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The integral on focus here is:

$$\int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx$$

I know a solution to this by considering a function $$f(a) = \int_0^1 \frac{\sin(\ln(ax))}{\ln x}\,dx$$ and then differentiating it w.r.t $a$ solving integral and then integrating again w.r.t. $a$.

Although I was wondering if you could perform this integration by any other method?

Mention the method in detail, please. :-)

Quanto
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Saket Gurjar
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6 Answers6

13

Substitute $\ln x = -t$,

\begin{align} \int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx =& \int_0^\infty \frac{e^{-t}}t \sin t \ dt =\int_0^\infty \int_1^\infty \sin t \ e^{-xt}dx \ dt\\ = &\int_1^\infty \frac1{1+x^2}dx =\frac\pi4\\ \end{align}

Quanto
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    Once more, you type too fast for the old man ! Cheers :-) – Claude Leibovici Apr 09 '20 at 03:16
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    This strategy proves more generally that$$\Re s>0\implies\int_0^1\frac{x^{s-1}\sin\ln x}{\ln x}dx=\int_0^\infty\frac{e^{-st}}{t}\sin tdt=\frac{\pi}{2}-\arctan s.$$ – J.G. Apr 09 '20 at 13:54
  • @J.G. - it’s a very pretty generalization – Quanto Apr 09 '20 at 14:14
  • I'm not an expert in calculus, so is this a standard approach in integration questions? i.e. writing $\frac{e^{-t}}{t}$ or rather any such term in terms of an definite integral involving another variable. Also I would really appreciate it if you could mention some sources to such approaches....... – Saket Gurjar Apr 09 '20 at 17:40
  • @SaketGurjar - certain integrals such as this one stem from double integrals. Switching the order of integration as done in the answer is a very effective technique – Quanto Apr 09 '20 at 18:41
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As in the post by @quanto, we begin by enforcing the substitution $x\mapsto e^{-x}$ to reveal

$$\int_0^1 \frac{\sin(\log(x))}{\log(x)}\,dx=\int_0^\infty \frac{e^{-x}\sin(x)}{x}\,dx\tag1$$

Next, we will use the identity from the Laplace Transform (See This)

$$\int_0^\infty f(x)g(x)\,dx=\int_0^\infty \mathscr{L}\{f\}(s)\,\,\mathscr{L}^{-1}\{g\}(s)\,ds\tag2$$

Let $f(x)=e^{-x}\sin(x)$ and $g(x)=\frac1x$. Then, we see that

$$\mathscr{L}\{f\}(s)=\frac{1}{(s+1)^2+1}\tag3$$

and

$$\mathscr{L}^{-1}\{g\}(s)=1\tag4$$

Hence, using $(3)$ and $(4)$ in $(2)$ and applying to $(1)$ yields

$$\begin{align} \underbrace{\int_0^\infty \frac{e^{-x}\sin(x)}{x}\,dx}_{\int_0^\infty f(x)g(x)\,dx}&=\underbrace{\int_0^\infty \frac{1}{(s+1)^2+1}\,ds}_{\int_0^\infty \mathscr{L}\{f\}(s)\,\,\mathscr{L}^{-1}\{g\}(s)\,ds}\\\\ &=\frac{\pi}{4} \end{align}$$

Integreek
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Mark Viola
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\begin{align}J&=\int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx\\ &=\int_0^1 \left(\sum_{n=0}^\infty\frac{(-1)^n\ln^{2n} x}{(2n+1)!}\right)\,dx\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\left(\int_0^1 \ln^{2n} x\,dx\right)\\ &=\sum_{n=0}^\infty\frac{(-1)^n(2n)!}{(2n+1)!}\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\\ &=\arctan(1)\\ &=\boxed{\frac{\pi}{4}}\\ \end{align}

NB: i assume $\displaystyle \int_0^1 \ln^{2n} x\,dx=(2n)!$ and $\displaystyle |x|\leq 1,\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$, and, $\displaystyle x\in\mathbb{R},\sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$

PS: $n\geq 0$, integer. \begin{align}K_n&=\int_0^1\ln^{n} x\,dx\\ K_0&=\int_0^1\ln^{0} x\,dx\\ &=1\\ K_1&=\Big[x(\ln x-1)\Big]_0^1\\ &=-1\\ K_{n+2}&=\Big[x(\ln x-1)\ln^{n+1} x\Big]_0^1-(n+1)\int_0^1 (\ln x-1)\ln^{n} x\,dx\\ &=(n+1)K_{n}-(n+1)K_{n+1}\\ \end{align} Proof by induction:

For $\displaystyle 0\leq m\leq n+1, K_m=(-1)^m\times m!$ \begin{align}K_{n+2}&=(n+1)K_{n}-(n+1)K_{n+1}\\ &=(n+1)\times(-1)^{n}\times n!-(n+1)\times (-1)^{n+1}\times (n+1)! \\ &=(-1)^n(n+2)!\left(\frac{1}{n+2}+\frac{n+1}{n+2}\right)\\ &=(-1)^n(n+2)!\\ &=\boxed{(-1)^{n+2}(n+2)!} \end{align}

FDP
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Using Feynman’s Integration Technique, we first let $$ I(a)=\int_0^1 \frac{\sin (a\ln x)}{\ln x} d x $$ Then differentiating $I(a)$ w.r.t. $a$ yields $$ I^{\prime}(a)=\int_0^1 \cos (a \ln x) d x $$ Letting $y=-a\ln x$ transforms

$$ I^{\prime}(a) =\frac{1}{a} \int_0^{\infty} e^{\frac{y}{a}} \cos y d y =\frac{1}{a^2+1} $$ Integrating back from $a=0$ to $1$ gives $$ \begin{aligned} \int_0^1 \frac{\sin (\ln x)}{\ln x} d x & =I(1)-I(0) \\ & =\int_0^1 \frac{1}{a^2+1} d a \\ & =\left[\tan ^{-1} a\right]_0^1 \\ & =\frac{\pi}{4} \end{aligned} $$

Lai
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$$I=\int_0^1\frac{\sin{\ln x}}{\ln x}dx=\Im\int_0^1\frac{x^i-1}{\ln x}dx$$ This is a classic integral to solve with Feynman's trick $$I(a)=\int_0^1\frac{x^a-1}{\ln x}dx, I(0)=0$$ $$I'(a)=\int_0^1\frac{x^a\ln x}{\ln x}dx=\int_0^1x^adx=\frac{x^{a+1}}{a+1}\Bigg|_0^1=\frac{1}{a+1}$$ $$I(a)=\ln(a+1)+C$$ $$I(0)=\ln(1)+C=C=0$$ $$I(a)=\ln(a+1)$$ Going back to $I$ $$I=\Im\int_0^1\frac{x^i-1}{\ln x}dx=\Im [I(i)]=\Im[\ln(1+i)]=\frac{\pi}{4}$$ We also derive another integral if we use the real part instead of the imaginary part $$J=\int_0^1\frac{\cos{\ln{x}}-1}{\ln x}dx=\Re[I(i)]=\frac{\ln 2}{2}$$

polychroma
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$$ \begin{aligned} \int_0^1 \frac{\sin (\ln x)}{\ln x} d x = & \Im\int_0^1 \frac{e^{i \ln x}}{\ln x} d x \\ = & \Im \int_0^1 \frac{x^i}{\ln x} d x \\ = & \Im \int_0^1 x^i \int_0^{\infty} x^{-y} d y d x \\ = & \Im\left[\int_0^{\infty} \int_0^1 x^{i-y} d x d y\right] \\ = & \Im\int_0^{\infty}\left[\frac{x^{i-y+1}}{i-y+1}\right]_0^1 d y \\ = & \Im \int_0^{\infty} \frac{1}{i-y+1} d y \\ =& -\int_0^{\infty} \frac{1}{(1-y)^2+1} d y\\=& \frac{\pi}{4} \end{aligned} $$

Lai
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