Log integrals I In this example the value of the integral I3=∫01ln3(1+x)xdx was derived. The purpose of this question is to determine the value of the more general form of the integral, namely, In=∫01lnn(1+x)xdx. Also is there a corresponding value for the integral Jn=∫01ln⁡n(1+x)1+xdx?

Question

Log integrals I  In this example the value of the integral  I3=∫01ln3(1+x)xdx  was derived. The purpose of this question is to determine the value of the more general form of the integral, namely,  In=∫01lnn(1+x)xdx.  Also is there a corresponding value for the integral  Jn=∫01ln⁡n(1+x)1+xdx?

Prince Huang · Accepted Answer

Derivation:  In=∫12lnnxx−1dx  =(−1)n∫121lnnxx−x2dx  =lnn+1(2)n+1+(−1)n∑k=1∞∫121xk−1lnndx  =lnn+1(2)n+1+(−1)n∑k=1∞∂n∂kn[1k−1k2k] (1)  =lnn+1(2)n+1+(−1)n (2)  ∑k=1∞[(−1)∩!kn+1−∑j=0n(nj)(−1)jj!kj+1(−1)n−jlnn−j(2)2k] (3)  =lnn+1(2)n+1+n!∑k=1n[1kn+1−∑j=0nlnn−j(2)kj+12k(n−k)!] (4)  =lnn+1(2)n+1+n![ζ(n+1)−∑j=0nln⁡n−j(2)(

ataill0k · Accepted Answer

As M.N.C.E. has provided a nice answer for In the only change that could be made it to cast the result into the form  In=∫01lnn(1+x)xdx=−nn+1lnn+1(2)+n!ζ(n+1)−∑k=1nn!lnn−k(2)(n−k)!Lik+1(12).  The second integral is  Jn=∫01lnn(1+x)1+xdx=[lnn+1(1+x)n+1]01=lnn+1(2)n+1

Log integrals I In this example the value of the integral I_3=\int^1_0\frac{\ln^3(1+x)}{x}dx was

Answered question

Answer & Explanation

New Questions in Algebra II