0
$\begingroup$

Question:-Prove that $\int_{0}^{1} \frac{\arccos x}{\ln x}dx=\sum_{k=0}^{\infty}-\frac {(2k-1)!!\ln(2k+2)}{2^k k! (2k+1)}$.

I have no idea how to prove this, If we use Taylor series of both $\arccos x$ and $\ln x $, then it is difficult to combine them. Also, Taylor series of $\frac {\arccos x}{\ln x}$ gives no help.

Can anybody help me to prove this?

$\endgroup$
3
$\begingroup$

We have : $$\arccos x=\frac{\pi}{2}-\arcsin x=\arcsin 1-\arcsin x$$ And using Taylor series of $\arcsin$ we get : $$\arccos x=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!(2n+1)}(1-x^{2n+1})$$ Then we get : $$\int_0^1\frac{\arccos x}{\ln x}=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!(2n+1)}\biggl(\int_0^1\frac{1-x^{2n+1}}{\ln x}\biggr)$$ Therefore we put $x=e^{-t}$ so we have By Frullani Theorem : $$\int_0^1\frac{1-x^{2n+1}}{\ln x}=\int_0^\infty\frac{e^{-t}-e^{-(2n+2)t}}{-t}=-\ln (2n+2) $$ Then we get the result.

$\endgroup$

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.