How do I evaluate the integral $\int_0^1 \frac{\sin ^n(\ln x)}{\ln ^m x}\,\mathrm d x$?

Question

Attracted by the decent answer of the integral in the post $$ \int_0^1 \frac{\sin (\ln x)}{\ln x}\,\mathrm d x=\frac{\pi}{4}, $$ I want to go further with $$ I_2=\int_0^1 \frac{\sin ^2(\ln x)}{\ln ^2 x}\,\mathrm d x=I(1)$$ where $$ I(a)=\int_0^1 \frac{\sin ^2(a \ln x)}{\ln ^2 x}\,\mathrm d x $$ whose first and second derivatives w.r.t. $a$ are

$$ \begin{aligned} I^{\prime}(a) & =\int_0^1 \frac{2 \sin (a \ln x) \cos (a \ln x)}{\ln x}\,\mathrm d x \\ & =\int_0^1 \frac{\sin (2 a \ln x)}{\ln x}\,\mathrm d x \\ I^{\prime \prime}(a) & =2 \int_0^1 \cos (2 a \ln x)\,\mathrm d x \\ & =2 \Re \int_0^1 e^{2i a \ln x}\,\mathrm d x \\ & =2 \Re \int_0^1 x^{2 a i}\,\mathrm d x \\ & =2 \Re\left(\frac{1}{2 a i+1}\right) \\ & =\frac{2}{1+4 a^2} \end{aligned} $$ Integrating back $$ \begin{aligned} I^{\prime}(a) & = I^{\prime}(a)-I(0)\\&=\int_0^a \frac{2}{1+4 x^2}\,\mathrm d x\\& =\tan ^{-1}(2 a) \end{aligned} $$ Integrating once more yields

$$ \begin{aligned} I_2 & =I(1)-I(0) \\ & =\int_0^1 I^{\prime}(a)\,\mathrm d a \\ & =\int_0^1 \tan ^{-1}(2 a)\,\mathrm d a \\ & =\left[a \tan ^{-1}(2 a)\right]_0^1-\int_0^1 \frac{2 a}{1+4 a^2}\,\mathrm d a \\ & =\tan ^{-1} 2-\frac{1}{4}\left[\ln \left(1+4 a^2\right)\right]_0^1 \\ & =\tan ^{-1} 2-\frac{1}{4} \ln 5 \end{aligned} $$

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My question

How do I tackle the integral $\int_0^1 \frac{\sin ^m(\ln x)}{\ln ^n x}\,\mathrm d x$?

Your comments and alternatives are highly appreciated.

Answer 1

Using: $$\frac{1}{\log ^2(x)}=\int_0^{\infty } x^{-t} t \, dt$$ and then:

$$\int_0^1 \frac{\sin ^2(\log (x))}{\log ^2(x)} \,\mathrm dx=\int_0^1 \sin ^2(\log (x)) \int_0^{\infty } x^t t \,\mathrm dt \,\mathrm dx=\int_0^{\infty } \left(\int_0^1 \sin ^2(\log (x)) x^t t \,\mathrm dx\right) \,\mathrm dt\\=\int_0^{\infty } \frac{2 t}{5+7 t+3 t^2+t^3} \,\mathrm dt=\frac{\pi }{2}-\cot ^{-1}(2)-\frac{\log (5)}{4}=\tan ^{-1}(2)-\frac{\log (5)}{4}$$

Generalization: (using: AI Gemini 2.5 Pro Preview 05-06)

$$\int_0^1 \frac{\sin ^n(\log (x))}{\log ^m(x)} \,\mathrm dx=\\\int_0^{\infty } (-1)^{m+n} e^{-x} x^{-m} \sin ^n(x) \, dx=\\\frac{(-1)^{m+n} \left((-1)^m (-i)^n\right) \sum _{k=0}^n \binom{n}{k} (-1)^k (1-i (n-2 k))^{m-1} \log (1-i (n-2 k))}{(m-1)! 2^n}$$

for: $n\geq m$

where: $\{n,m\}\in \mathbb{Z}_{>\, 0}$

UPDATE:

Mathematica code:

With: n=1..100 and m=1..100

  f[n_, m_] /; n >= m := Chop[N[(-1)^(m + n)*(((-1)^m*(-I)^n)/(Factorial[m - 1]*2^n))*
  Sum[Binomial[n, k]*(-1)^k*(1 - I*(n - 2*k))^(m - 1)*
  Log[1 - I*(n - 2*k)], {k, 0, n}]]]; 
  g[n_, m_] /; n >= m := Chop[NIntegrate[(-1)^(m + n)*Exp[-x] x^(-m)*Sin[x]^n, {x, 0, Infinity},Method -> "LocalAdaptive", WorkingPrecision -> 30, MaxRecursion -> 100]]; 
  CompareNumerically[val1_, val2_, places_Integer] := 
  Round[val1, 10.0^(-places)] == Round[val2, 10.0^(-places)]; 
  Table[CompareNumerically[f[n, m], g[n, m], 5], {n, 1, 100}, {m, 1, n}] // Flatten
(* True,True,...True *)

Answer 2

Substitute $\ln x = -t$, along with $g_n(t)=e^{-t}\sin^n t$

\begin{align} I_n=&\int_0^1\frac{\sin^n(\ln x)}{\ln^n x}\,\mathrm dx\\ =&\int_0^\infty \frac{g_n(t)}{t^n}\,\mathrm dt \overset{\text{ibp}} =\frac1{(n-1)!}\int_0^\infty \frac{g_n^{(n-1)}(t)}t\,\mathrm dt\\ =&\ \frac1{(n-1)!}\int_0^\infty \int_0^\infty g_n^{(n-1)}e^{-st}\,\mathrm ds\ \overset{\text{ibp}}{\mathrm dt}\\ =&\ \frac1{(n-1)!}\int_0^\infty \int_0^\infty s^{n-1}e^{-st}g_n(t)\,\mathrm dt\,\mathrm ds\\ =&\ \frac1{(n-1)!}\int_0^\infty s^{n-1}h_n(s)\,\mathrm ds \end{align} where $$h_n(s)= \int_0^\infty e^{-(s+1)t}\sin^n t\,\mathrm dt =\frac{n(n-1)}{(s+1)^2+n^2}h_{n-2}(s) $$ Then, with $h_0(s)=\frac1{s+1}$ and $h_1(s)=\frac1{(s+1)^2+1}$ \begin{align} I_1=&\int_0^\infty \frac{1}{(s+1)^2+1}\,\mathrm ds=\frac\pi4\\ I_2=&\int_0^\infty \frac{2s}{(1+s)[(s+1)^2+4]}\,\mathrm ds=\tan^{-1}2-\frac14\ln5\\ I_3=&\int_0^\infty \frac{3s^2}{[(s+1)^2+1] [(s+1)^2+9]}\,\mathrm ds=\tan^{-1}3-\frac38\ln5\\ I_4=&\int_0^\infty \frac{4s^3}{(s+1)[(s+1)^2+4] [(s+1)^2+16]}\,\mathrm ds\\ = &\ \frac{13}{12}\tan^{-1}4-\frac{1}{6}\tan^{-1}2 +\frac{11}{24}\ln5-\frac{47}{96}\ln(17)\\ I_5=&\int_0^\infty \frac{5s^4}{[(s+1)^2+1] [(s+1)^2+9] [(s+1)^2+25]}\,\mathrm ds\\ = &\ \frac{119}{96}\tan^{-1}5-\frac{35}{96}\tan^{-1}3 -\frac{5}{8}\ln\frac{13}5-\frac{5\pi}{192}\\ I_6=&\cdots & \end{align}

Answer 3

Let $t=-\ln(x)$ \begin{align*} I &= \int^{\infty}_0 \frac{\sin^2(t)}{t^2}e^{-t}\,\mathrm dt \\ &= \lim\limits_{s\to 1} \mathcal{L} \left({\frac{\sin^2(t)}{t^2}}\right) \\ &= \arctan(2)-\frac{1}{4}\ln(5) \end{align*} More generally I don't know if there is a closed form in this case.

\begin{align*} I &= \int^{\infty}_0 \frac{\sin^n(t)}{t^n}e^{-t}\,\mathrm dt \\ &= \lim\limits_{s\to 1} \mathcal{L} \left({\frac{\sin^n(t)}{t^n}}\right) \end{align*}

Answer 4

Triple integral

Let $y=-\ln x$, then $$ \begin{aligned} I_2 & =\frac{1}{2} \int_0^{\infty} \frac{e^{-y}}{y} \cdot \frac{1-\cos 2 y}{y}\,\mathrm d y \\ & =\frac{1}{2} \int_0^{\infty} \int_1^{\infty} e^{-v y} d v \int_0^2 \sin (u y)\,\mathrm d u\,\mathrm d y \\ & =\frac{1}{2} \int_0^2 \int_1^{\infty} \int_0^{\infty} e^{-v y} \sin (u y)\,\mathrm d y\,\mathrm d v\,\mathrm d u\\ & =\frac{1}{2} \int_0^2 \int_1^{\infty} \frac{u}{u^2+v^2}\,\mathrm d v\,\mathrm d u \\ & =\frac{1}{2} \int_0^2\tan ^{-1}u\,\mathrm d u \\ &= \tan^{-1} 2-\frac{1}{4}\ln 5 \end{aligned} $$