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How do we evaluate

$$I=\int_{0}^{1}\frac{\sin[a\ln(1-x)]}{\ln(1-x)}\cdot x\mathrm dx$$

Making a sub: $u=\ln(1-x)$ this lead to a messy integral.

I believe the closed form for $I=\arctan\left(\frac{a}{a^2+2}\right)$, how to shows it, I am not sure. We leaves it to the expert!

3 Answers3

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Let $$I(a)=\int_{0}^{1}\frac{\sin[a\ln(1-x)]}{\ln(1-x)}\cdot x\mathrm dx$$ Then $$\begin{split} I^\prime(a)&=\int_0^1x\cos[a\ln(1-x)]dx\\ &=\int_{0}^{+\infty}(1-e^{-t})\cos(-at)e^{-t}dt \,\,\,\left(\text{via }t=-\ln(1-x)\right)\\ &=\int_0^{+\infty}(e^{-t}-e^{-2t})\frac{e^{iat}+e^{-iat}}2dt\\ &=\int_0^{+\infty}\frac{e^{(ia-1)t}+e^{-(ia+1)t}-e^{(ia-2)t}-e^{-(ia+2)t}}{2}dt\\ &=\frac 1 2 \left(\frac{1}{1-ia} + \frac{1}{1+ia} -\frac{1}{2-ia} - \frac{1}{2+ia}\right)\\ &=\frac 1 {1+a^2}-\frac{2}{a^2+4} \end{split}$$ Note that $I(0)=0$ and integrate: $$I(a)=\arctan(a)-\arctan(\frac a 2)$$ which can be simplified into $$\boxed{I(a)=\arctan\left(\frac{a}{a^2+2}\right)}$$

Stefan Lafon
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By Feynman’s Integration Technique

First of all, letting $ x\mapsto 1-x$ transforms the integral into $$ I(a)=\int_0^1 \frac{\sin (a \ln x)}{\ln x}(1-x) d x $$ Differentiating $I(a)$ w.r.t. $a$ yields $$ I^{\prime}(a)=\int_0^1(1-x) \cos (a \ln x) d x $$ Putting $y=-\ln x$ gives $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} e^{-y}\left(1-e^{-y}\right) \cos (a y) d y \\ & =\int_0^{\infty} e^{-y} \cos (a y) d y-\int_0^{\infty} e^{-2 y} \cos (a y) d y \\ & =\frac{1}{a^2+1}-\frac{2}{a^2+4} \end{aligned} $$

Integrating back, we have $$ \begin{aligned} \int_0^1 \frac{\sin [a \ln (1-x)]}{\ln (1-x)} \cdot x d x & =\int_0^a\left(\frac{1}{x^2+1}-\frac{2}{x^2+4}\right) d x \\ & =\tan ^{-1} a-\tan ^{-1}\left(\frac{a}{2}\right) \\ & =\tan ^{-1}\left(\frac{a}{a^2+2}\right) \end{aligned} $$

Lai
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By Double integral

Letting $1-x=e^{-y}$ transforms the integral into $$ \begin{aligned} I & =-\int_0^{\infty} \frac{e^{-y}-e^{-2 y}}{y} \sin (a y) d y \\ & =\int_0^{\infty}\left[\int_1^2 e^{-t y} d t \cdot \sin (a y)\right] d y \\ & =\int_1^2\left[\int_0^{\infty} e^{-t y} \sin (a y) d y\right] d t \\ & =\int_1^2 \frac{a}{a^2+t^2} d t \\ & =\left[\tan ^{-1}\left(\frac{t}{a}\right)\right]_1^2 \\ & =\tan ^{-1} \frac{2}{a}-\tan ^{-1} \frac{1}{a} \\ & =\tan ^{-1}\left(\frac{a}{a^2+2}\right) \end{aligned} $$

Lai
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