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Integral of sin(ln^2(x))/x dx

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The solution

You have entered [src]
  1                
  /                
 |                 
 |     /   2   \   
 |  sin\log (x)/   
 |  ------------ dx
 |       x         
 |                 
/                  
0                  
$$\int\limits_{0}^{1} \frac{\sin{\left(\log{\left(x \right)}^{2} \right)}}{x}\, dx$$
Integral(sin(log(x)^2)/x, (x, 0, 1))
The answer (Indefinite) [src]
  /                        /               
 |                        |                
 |    /   2   \           |    /   2   \   
 | sin\log (x)/           | sin\log (x)/   
 | ------------ dx = C +  | ------------ dx
 |      x                 |      x         
 |                        |                
/                        /                 
$$\int \frac{\sin{\left(\log{\left(x \right)}^{2} \right)}}{x}\, dx = C + \int \frac{\sin{\left(\log{\left(x \right)}^{2} \right)}}{x}\, dx$$
The answer [src]
  1                
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 |                 
 |     /   2   \   
 |  sin\log (x)/   
 |  ------------ dx
 |       x         
 |                 
/                  
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$$-\lim_{x\downarrow 0}{{{\sqrt{\pi}\,\left(\sqrt{2}\,i+\sqrt{2} \right)\,\mathrm{erf}\left({{\left(\sqrt{2}\,i+\sqrt{2}\right)\, \log x}\over{2}}\right)}\over{16}}+{{\sqrt{\pi}\,\left(\sqrt{2}\,i- \sqrt{2}\right)\,\mathrm{erf}\left({{\left(\sqrt{2}\,i-\sqrt{2} \right)\,\log x}\over{2}}\right)}\over{16}}+{{\sqrt{\pi}\,\left( \sqrt{2}-\sqrt{2}\,i\right)\,\mathrm{erf}\left(\sqrt{-i}\,\log x \right)}\over{16}}+{{\sqrt{\pi}\,\left(\sqrt{2}\,i+\sqrt{2}\right)\, \mathrm{erf}\left(\left(-1\right)^{{{1}\over{4}}}\,\log x\right) }\over{16}}}$$
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  1                
  /                
 |                 
 |     /   2   \   
 |  sin\log (x)/   
 |  ------------ dx
 |       x         
 |                 
/                  
0                  
$$\int\limits_{0}^{1} \frac{\sin{\left(\log{\left(x \right)}^{2} \right)}}{x}\, dx$$
Numerical answer [src]
0.0235419702091351
0.0235419702091351

    Use the examples entering the upper and lower limits of integration.