0
$\begingroup$

Finding $\displaystyle \lim_{a\rightarrow\infty}\int^{1}_{0}\frac{\arctan(ax)\cdot \ln(1+x)}{1+x^2}dx$

What I try put $x=\tan \theta\,$ and $\,dx=\sec^2\theta\, d\theta$

$$I=\lim_{a\rightarrow\infty}\int^{\frac{\pi}{4}}_{0}\arctan(a\cdot \tan\theta)\cdot \ln(1+\tan\theta)\,d\theta$$

How do I solve it? Help me please.

$\endgroup$
  • 1
    $\begingroup$ Apply DCT then the given integral will be $\frac{\pi}{2} \int_0^1 \frac{\ln(x+1)}{x^2 + 1}dx$ and you can evaluate this with $x=\tan u $ easily. $\endgroup$ – bFur4list Jan 27 at 20:14
  • 1
    $\begingroup$ After dominated convergence, you get what's known as Serret's integral. It's solved here: math.stackexchange.com/questions/155941/… $\endgroup$ – bjorn93 Jan 28 at 0:34
1
$\begingroup$

Since $\frac{\log(1+x)}{1+x^2}$ is continuous in $[0,1]$ then there exists $K \in \mathbb{R}$ such that $|\frac{log(1+x)}{1+x^2}| \leq K, \forall x \in [0,1]$. Since $\arctan$ is a bounded function you will get $ \int (*) \leq K.\frac{\pi}{2}$ so you get a convergent integral.

| 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.