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Wolfram Alpha thinks that it is unsolvable, but...

My solution so far:

$$\int \frac{\sin(\pi x) }{\ln x}dx = \int \frac{e^{i\pi x} -e^{-i\pi x}}{2i\ln x}dx.$$

Let us split sum and focus on the first integral, using the Feynman's technique:

$$\int \frac{e^{i\pi x}}{\ln x}dx = I(0), $$

$$ I(t)= \int \frac{e^{i\pi x}x^{-t}}{\ln x}dx, $$

$$ I'(t)= \int \frac{e^{i\pi x}x^{-t}(-\ln x)}{\ln x}dx = - \int e^{i\pi x}x^{-t}dx.$$

Substitute $-i\pi x = u$, $-i\pi dx = du$, $dx = \frac{du}{-i\pi}$,

$$ I'(t)= \int e^{-u}\left(\frac{u}{-i\pi}\right)^{-t}\frac{du}{-i\pi}= \frac{1}{(-i\pi)^{-t+1}}\int e^{-u}u^{-t}du = \frac{1}{(-i\pi)^{1-t}} \Gamma (1-t).$$

Now I must integrate it with respect to $t$.

Any ideas how to solve it further or maybe you can suggest other method?

Gary
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user184868
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    Feynman's trick is applicable to definite integrals, not to anti-derivatives. – Mark Viola Feb 02 '22 at 22:59
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    In any case, you end up having to integrate an expression that doesn't look much simpler than the initial one... – Tob Ernack Feb 02 '22 at 23:00
  • These integrals tend to only be evaluated on improper definite bounds (still a difficult job in itself); finding an antiderivative of a non elementary function like this will give you a nasty series, or a whole host of special functions, or nothing - what is your motivation for a “closed” form? – FShrike Feb 02 '22 at 23:04
  • @Tob Ernack important is not the simplicity, but what can i do with the expresion. any thought on this gamma, if it stays under the integral? – user184868 Feb 02 '22 at 23:07
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    @FShrike finding the solution, no one found before is the motivation :D – user184868 Feb 02 '22 at 23:08
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    The lack of definite bounds causes problems. You wrote $\cdots=\Gamma(1-t)$ but that equality would be true only if the bounds were $0,\infty$ – FShrike Feb 02 '22 at 23:10
  • @FShrike oh thanks i lot let the bounds be like that first :D. So can i now substitute t=0 in the I'(t)? – user184868 Feb 02 '22 at 23:12
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    Wolfram alpha probably knows all these tricks. If it cannot find an antiderivative, you cannot too. – markvs Feb 02 '22 at 23:13
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    @markvs For the OP’s benefit I’ll say that on the whole, this site can be better than wolfram at evaluating definite integrals, but I agree that an indefinite integral cannot be found by a human if it cannot be found by a computer library. – FShrike Feb 02 '22 at 23:15
  • I am not sure even about definite integrals. It is like playing chess with Stockfish. – markvs Feb 02 '22 at 23:20

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