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I was wondering how to prove this Lambert W identity.

$$ - \pi W(-1) = \int_{-\infty}^{\frac{-1}{e}} \Im ( W ' (x) ) \ln(1 + \frac{1}{x} ) dx $$

Maybe with contour integration ??

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    $'$ means derivative ofcourse. And $Im$ imaginary part – 2017-02-09
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    I guess that the identity $W'(x)=\frac{W(x)}{x(1+W(x))}$ for the principal branch of the Lambert function and the substitution $x=te^t$ turn the integral into a simple one. – 2017-02-09
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    Also, integration by parts leads to the integral of the imaginary part of $\frac{W(x)}{x+x^2}$. – 2017-02-09
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    we are talking about the zeroth branch of $W$? – 2017-02-09
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    @JackD'Aurizio using a contour integration approach on your second proposed form of the integrand yields indeed the conjectured result (at least if one defines things much more accurate then op did) – 2017-02-09
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    @OP do you mean $W(-1+i\delta)$ or $W(-1-i\delta)$ – 2017-02-09
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    How do we get rid of the imaginary part piece in the integrand ? – 2017-02-10
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    any reason why youa aren't answering my questions? – 2017-02-10
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    as this stands your original question is highly ill defined – 2017-02-10
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    Im talking about the standard definition of the Lambert W function ! How can a standard be confusing ? Any other interpretation makes the statement untrue anyways ? – 2017-02-10
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    do you know what a branch cut is? – 2017-02-13
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    Yes like the ln has a cut on the negative reals. That does not answer the OP. Im using standard def here , so im not sure what you are suggesting here. Post an answer or give a better hint. – 2017-02-15

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