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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.

Bernard
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jacky
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    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 '20 at 20:14
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    After dominated convergence, you get what's known as Serret's integral. It's solved here: https://math.stackexchange.com/questions/155941/evaluate-the-integral-int-01-frac-lnx1x21-mathrm-dx – bjorn93 Jan 28 '20 at 00:34

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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.

Bernard
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astro
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