Leibniz Integral Rule

Author

Eric W. Weisstein
October 22, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Calculus/LeibnizIntegralRule.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/LeibnizIntegralRule.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Example

(∫_0^π)ln(-2 α cos(x)+(α^2)+1)dx==2 π ln(α)

V5.0.1

Wrong (38933)

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α<-1]

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->-1<α<1]

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α>1]

V6

$Version

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α>1]//Timing

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α∈Reals]//Timing

...useless...

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi}]//Timing

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α<-1||α>1]

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α<-1]

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->-1<α<1]

FullSimplify[π (-Log[2]+Log[1+(α^2)]+Log[1+(Sqrt[1-((4 (α^2))/(((1+(α^2)))^2))])]),-1<α<1]

Table[NIntegrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},AccuracyGoal->Infinity,MaxRecursion->20],{α,-.95,.95,.1}]//Chop//Union

V7

$Version

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α>1]//Timing

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi},Assumptions->α∈Reals]//Timing

...wrong...

Integrate[Log[1-2α Cos[x]+α^2],{x,0,Pi}]//Timing

FullSimplify[Last[%],α>1]

FullSimplify[(1/2) π (I π-2 Log[-1-α]+2 Log[-α]-2 Log[1+α]+2 Log[(((1+α))^2)])+I PolyLog[2,-(1/α)]+I PolyLog[2,-α]-I (PolyLog[2,(1/α)]+PolyLog[2,α]),α>1]