Prove that ∫ 0 1 ln ⁡ ( x ) x − 1 d x = ∑ k = 1 ∞ 1 k 2

Question

Prove that       ∫          0              1                  ln      ⁡      (      x      )              x      −      1        d  x  =      ∑          k      =      1              ∞            1          k      2

zlepljalz2 · Accepted Answer

When applying the Weierstrass M-test, use the fact that      ∑          k      =      1              ∞            |    −                  (        −        1                  )                      k            −            1                          (        β        −        1                  )                      k                                      k        2              |    ≤      ∑          k      =      1              ∞            1          k      2        &amp;lt;  ∞so that the sum is dominated by a sum that&#039;s independent of   β. Then the limit as   β  →      0    +   may proceed.

ddcon4r · Accepted Answer

By definition      ∑          k      =      1              ∞        (  −  1      )          k                  (      β      −      1              )                  k                            k      2        =      Li    2    (  1  −  β  )If   β is small      Li    2    (  1  −  β  )  =  ζ  (  2  )  +  β  log  ⁡      (          β      e        )    +  O      (          β      2        )

Prove that <msubsup> ∫ <mrow class="MJX-TeXAtom-ORD"> 0 </mrow

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