5
$\begingroup$

In formula 4.224 12 in the 8. edition of the Table of Integrals, Series, and Products is an error: In the case $a^2=1$ both equations should hold, but they are not equal. I think the second equation (i.e. $\int_0^\pi \ln(1+a\cos x)dx = 2 \pi \ln \frac{|a|}{2}$ ($a^2 \geq 1$) is wrong and it should be $$\int_0^\pi \ln((1+a\cos x)^2)dx = 2 \pi \ln \frac{|a|}{2}$$ This would also solve the problem, that the argument of the logarithm gets negative if $a^2 \geq 1$. Can someone confirm this and/or does someone know how to prove this?

$\endgroup$
5
$\begingroup$

this is entry 172 in the 2005 Errata, which somehow didn't make it properly into the 8th edition.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.