Show that $\lim_{n\rightarrow\infty}\int_{0}^{n}\frac{\sqrt x\ln x}{(1+x)^2}\,dx=\pi$ Show that $\lim_{n\rightarrow\infty}\int_{0}^{n}\frac{\sqrt x\ln x}{(1+x)^2}\,dx=\pi$. This limit puzzled me as I never worked on this kind before and I am inclined to think that, as the $n$ is in the integral's boundary, there is a theorem involved like the dominant convergence theorem. I tried integrating by parts, but the $0$ of the integral prevented me from applying $log0$. How should I handle it?

The integrand is positive for $x>1$, so normally we'd just state the problem as $$\int_0^\infty\frac{\sqrt{x}\ln x}{(1+x)^2}dx=\pi.$$Let's first note that the substitution $x=\tan^2 t$ allows us to solve a seemingly unrelated problem, $$\int_0^\infty\frac{x^{k-1}}{(1+x)^2}dx=\int_0^{\pi/2}2\sin^{2k-1}t\cos^{3-2k} tdt=\operatorname{B}(k,\,2-k)\\=\Gamma(k)\Gamma(1-k)=\pi(1-k)\csc\pi k.$$(Look up Beta and Gamma functions if you don't know them well.) But this problem is not unrelated! Let's differentiate with respect to $k$: $$\int_0^\infty\frac{x^{k-1}\ln x}{(1+x)^2}dx=-\pi\csc\pi k[1+(1-k)\cot\pi k].$$Finally, substituting $k=\frac{3}{2}$ gives $$\int_0^\infty\frac{\sqrt{x}\ln x}{(1+x)^2}dx=\pi.$$
