# definite integration with limit approches to infinity

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.

• 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. – bFur4list Jan 27 at 20:14
• After dominated convergence, you get what's known as Serret's integral. It's solved here: math.stackexchange.com/questions/155941/… – bjorn93 Jan 28 at 0:34

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.