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Revision as of 03:58, 29 January 2024 by Martin.s (talk | contribs) (The blog is dedicated to advanced integrals sums and series. Hope you enjoy)
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\[\int_0^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}=2\underset{I}{\underbrace{\int_0^1{\frac{\ln ^2\left( \dfrac{2}{1+x} \right) \ln ^3\left( \dfrac{2x}{1+x} \right)}{\left( 1+x \right) ^2}\text{d}x}}}\]


ln2(21+x)ln3(2x1+x)=ln22ln3(1+x)+2ln2ln4(1+x)ln5(1+x)+3ln22ln2(1+x)ln(2x)6ln2ln3(1+x)ln(2x)+3ln4(1+x)ln(2x)3ln22ln(1+x)ln2(2x)+6ln2ln2(1+x)ln2(2x)3ln3(1+x)ln2(2x)+ln22ln3(2x)2ln2ln(1+x)ln3(2x)+ln2(1+x)ln3(2x)

\[I_1=-\ln ^22\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=\frac{\ln ^52}{2}+\frac{\text{3}\ln ^42}{2}+\text{3}\ln ^32-\text{3}\ln ^22\]

\[I_2=\text{2}\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-\ln ^52-\text{4}\ln ^42-\text{12}\ln ^32-\text{24}\ln ^22+\text{24}\ln 2\]

\[I_3=-\int_0^1{\frac{\ln ^5\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-60+\frac{\ln ^52}{2}+\frac{\text{5}\ln ^42}{2}+\text{10}\ln ^32+\text{30}\ln ^22+\text{60}\ln 2\]

I4=3ln2201ln2(1+x)ln(2x)(1+x)2dx=3ln2201ln2(1+x)lnx(1+x)2dx+3ln3201ln2(1+x)(1+x)2dx

\[I_{41}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2\]

\[I_{42}=I_{82}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2\]

I5=6ln201ln3(1+x)ln(2x)(1+x)2dx=6ln201ln3(1+x)lnx(1+x)2dx6ln2201ln3(1+x)(1+x)2dx

I51=01ln3(1+x)lnx(1+x)2dx=01ln3(1+x)x(1+x)dx+301lnxln2(1+x)(1+x)2dx=01ln3(1+x)xdx01ln3(1+x)1+xdx+301lnxln2(1+x)(1+x)2dx=[6Li4(12)214ln2ζ(3)+115π414ln42+14π2ln22]ln424+3[14ζ(3)+π2613ln32ln222ln2]=6Li4(12)214ln2ζ(3)+115π412ln42ln32+14π2ln22+34ζ(3)+π223ln226ln2

\[I_{52}=\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=3-\frac{1}{2}\ln ^32-\frac{3}{2}\ln ^22-\text{3}\ln 2\]

\[I_6=3\int_0^1{\frac{\ln ^4\left( 1+x \right) \ln \left( 2x \right)}{\left( 1+x \right) ^2}\text{d}x}=3\int_0^1{\frac{\ln ^4\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x}+\text{3}\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}\]

I61=01ln4(1+x)lnx(1+x)2dx=01ln4(1+x)x(1+x)dx+401lnxln3(1+x)(1+x)2dx=01ln4(1+x)xdx01ln4(1+x)1+xdx+401lnxln3(1+x)(1+x)2dx=[24Li5(12)24ln2Li4(12)+24ζ(5)212ln22ζ(3)45ln52+23π2ln32]15ln52+4[6Li4(12)214ln2ζ(3)+115π412ln42ln32+14π2ln22+34ζ(3)+π223ln226ln2]=24Li5(12)24Li4(12)24ln2Li4(12)+24ζ(5)212ln22ζ(3)21ln2ζ(3)ln52+415π42ln424ln32+23π2ln32+π2ln22+3ζ(3)+2π212ln2224ln2

\[I_{62}=\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=12-\frac{1}{2}\ln ^42-\text{2}\ln ^32-\text{6}\ln ^22-\text{12}\ln 2\]

I7=3ln2201ln(1+x)ln2(2x)(1+x)2dx=3ln2201ln(1+x)ln2x(1+x)2dx3ln4201ln(1+x)(1+x)2dx6ln3201ln(1+x)lnx(1+x)2dx

I71=01ln(1+x)ln2x(1+x)2dx=01ln2x(1+x)2dx+201ln(1+x)lnxx(1+x)dx=01ln2x(1+x)2dx+201ln(1+x)lnxxdx201ln(1+x)lnx1+xdx=16π2+2[34ζ(3)]2[18ζ(3)]=54ζ(3)+16π2

\[I_{72}=\int_0^1{\frac{\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{2}-\frac{1}{2}\ln 2\]

\[I_{73}=\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{12}\pi ^2-\frac{1}{2}\ln ^22-\ln 2\]

I8=6ln201ln2(1+x)ln2(2x)(1+x)2dx=6ln201ln2(1+x)ln2x(1+x)2dx+6ln3201ln2(1+x)(1+x)2dx+12ln2201ln2(1+x)lnx(1+x)2dx

I81=01ln2(1+x)ln2x(1+x)2dx=201ln(1+x)ln2x(1+x)2dx+201ln2(1+x)lnxx(1+x)dx=201ln2x(1+x)2dx+401ln(1+x)lnxx(1+x)dx+201ln2(1+x)lnxxdx201ln2(1+x)lnx1+xdx=201ln2x(1+x)2dx+401ln(1+x)lnxxdx401ln(1+x)lnx1+xdx+201ln2(1+x)lnxxdx201ln2(1+x)lnx1+xdx=216π2+4[34ζ(3)]4[18ζ(3)]+2[4Li4(12)72ln2ζ(3)+124π416ln42+16π2ln22]2[2Li4(12)+74ln2ζ(3)145π4+112ln42112π2ln22]=12Li4(12)212ln2ζ(3)52ζ(3)+23180π412ln42+12π2ln22+13π2

\[I_{82}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2\]

I83=01ln2(1+x)lnx(1+x)2dx=01ln2(1+x)x(1+x)dx+201lnxln(1+x)(1+x)2dx=01ln2(1+x)xdx01ln2(1+x)1+xdx+201lnx(1+x)2dx+201ln(1+x)x(1+x)dx=01ln2(1+x)xdx01ln2(1+x)1+xdx+201lnx(1+x)2dx+201ln(1+x)xdx201ln(1+x)1+xdx=14ζ(3)ln323+2(ln2)+2π2122ln222=14ζ(3)+π2613ln32ln222ln2

I9=301ln3(1+x)ln2(2x)(1+x)2dx=301ln3(1+x)ln2x(1+x)2dx3ln2201ln3(1+x)(1+x)2dx3ln201ln3(1+x)lnx(1+x)2dx

I91=01ln3(1+x)ln2x(1+x)2dx=201lnxln3(1+x)x(1+x)dx+301ln2xln2(1+x)(1+x)2dx=201lnxln3(1+x)xdx201lnxln3(1+x)1+xdx+301ln2xln2(1+x)(1+x)2dx=2[12Li5(12)12ln2Li4(12)+12π2ζ(3)+9916ζ(5)25ln52+13π2ln32214ln22ζ(3)]2[6Li5(12)+6ln2Li4(12)6ζ(5)+218ln22ζ(3)+15ln5216π2ln32]+3[12Li4(12)212ln2ζ(3)52ζ(3)+23180π412ln42+12π2ln22+13π2]=36Li5(12)36Li4(12)36ln2Li4(12)+π2ζ(3)+1958ζ(5)152ζ(3)65ln52+π2ln32634ln22ζ(3)632ln2ζ(3)+2360π432ln42+32π2ln22+π2

\[I_{92}=I_{52}=\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=3-\frac{1}{2}\ln ^32-\frac{3}{2}\ln ^22-\text{3}\ln 2\]

I93=I51=01ln3(1+x)lnx(1+x)2dx=6Li4(12)214ln2ζ(3)+115π412ln42ln32+14π2ln22+34ζ(3)+π223ln226ln2

\[I_{10}=\ln ^22\int_0^1{\frac{\ln ^3\left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x=-\frac{9}{2}\ln ^22\zeta \left( 3 \right) -\frac{5}{2}\ln ^52+\frac{1}{2}\pi ^2\ln ^32\]

I11=2ln201ln(1+x)ln3(2x)(1+x)2dx=2ln4201ln(1+x)(1+x)2dx6ln3201ln(1+x)lnx(1+x)2dx6ln2201ln(1+x)ln2x(1+x)2dx2ln201ln(1+x)ln3x(1+x)2dx

\[I_{111}=I_{72}=\int_0^1{\frac{\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{2}-\frac{1}{2}\ln 2\]

\[I_{112}=I_{73}=\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{12}\pi ^2-\frac{1}{2}\ln ^22-\ln 2\]

\[I_{113}=I_{71}=\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x=-\frac{5}{4}\zeta \left( 3 \right) +\frac{1}{6}\pi ^2\]

I114=01ln(1+x)ln3x(1+x)2dx=01ln3x(1+x)2dx+301ln(1+x)ln2xx(1+x)dx=01ln3x(1+x)2dx+301ln(1+x)ln2xxdx301ln(1+x)ln2x1+xdx=92ζ(3)+37π43603[4Li4(12)+72ln2ζ(3)124π4+16ln4216π2ln22]=12Li4(12)212ln2ζ(3)92ζ(3)+1160π412ln42+12π2ln22

I12=01ln2(1+x)ln3(2x)(1+x)2dx=ln3201ln2(1+x)(1+x)2dx+3ln2201ln2(1+x)lnx(1+x)2dx+3ln201ln2(1+x)ln2x(1+x)2dx+01ln2(1+x)ln3x(1+x)2dx

\[I_{121}=I_{42}=I_{82}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2\]

\[I_{122}=I_{41}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2\]

\[I_{123}=I_{81}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}\text{d}x}=-\text{12Li}_4\left( \frac{1}{2} \right) -\frac{21}{2}\ln 2\zeta \left( 3 \right) -\frac{5}{2}\zeta \left( 3 \right) +\frac{23}{180}\pi ^4-\frac{1}{2}\ln ^42+\frac{1}{2}\pi ^2\ln ^22+\frac{1}{3}\pi ^2\]

I124=01ln2(1+x)ln3x(1+x)2dx=201ln(1+x)ln3x(1+x)2dx+301ln2(1+x)ln2xx(1+x)dx=201ln(1+x)ln3x(1+x)2dx+301ln2(1+x)ln2xxdx301ln2(1+x)ln2x1+xdx=2[12Li4(12)212ln2ζ(3)92ζ(3)+1160π412ln42+12π2ln22]+3[13π2ζ(3)295ζ(5)]3[8Li5(12)+8ln2Li4(12)338ζ(5)13π2ζ(3)+72ln22ζ(3)+415ln5229π2ln32]=24Li5(12)24Li4(12)24ln2Li4(12)20140ζ(5)212ln22ζ(3)21ln2ζ(3)+2π2ζ(3)9ζ(3)+1130π4ln42+π2ln2245ln52+23π2ln32