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HARD INTEGRAL - PART II

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       Today We will compute the following integral following the same ideas of the previous post : Recall (see here ) (1) Integrating both sides of (1)  from 0 to Computing    Recall (see here ) Letting we obtain We are looking for the Imaginary part of the equation above: Computing the quantities: The Glaisher function We know that (see here ): If we integrate from 0 to x we obtain Integrating from 0 to x we obtain

HARD INTEGRAL - PART I

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Today´s blog we will evaluate a hard integral, namely: Where is Catalan´s constant (G=0.915965594177219015054603514932384110774)  and is the Polylogarithm function . Recall the expansion (see here ) (1) Therefore (2) Integrating both sides of from 0 to We used that (see Appendix below): And Appendix Recall the generator function (see this post) (A.1) Letting in (A.1) we obtain We are looking for the Imaginary part of the equation above, hence Where we used the following results(see proofs here ): Recall (see this post ) (A.2) Integrating twice both sides of (A.2) from 0 to x we obtain: Letting we obtain:

Hjortnes series for zeta(3) - PART II

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We have previously showed that Today we will revisit this remarkable result proving it by a different method, namely, we will show that And compute this integral. First, lets find a series expansion for   : Letting we obtain (1) Now recall (see here ) (2) Letting we obtain (3) Claim: Proof: Letting k=3  we obtain Let´s now calculate this integral Where we used (see here )