$Proposition:$ $$\int_{0}^{\infty}\frac{1-\cos(x)}{x^{2}}dx=\frac{\pi}{2}$$ Initially the function:$f(z)=\frac{1-e^{iz}}{z^{2}}$ was considered due to the fact we have a trigonometric term within our numerator.Our function $f(z)$ was integrated on the upper semicircle within our Contour,our Se...

For the first one, write: $$\lim_{x\to 0^+} x\cdot \ln(x+x^2)=\lim_{x\to 0^+}x\cdot \ln(x(x+1))=\lim_{x\to 0^+}x(\ln(x)+\ln(x+1))$$ Hence, it remains to solve: $$\lim_{x\to 0^+}x\ln(x)+\lim_{x\to 0^+} x\ln(x+1)=\lim_{x\to 0^+} x\ln{x} \tag{1}$$ You can now apply L'Hopital's rule if you write: $$\...

By the ratio test, since we know that $$\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^{n/2}}{\sqrt{n+1}}\le\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^n}{\sqrt{n+1}}=\frac e\infty=0$$ Then it follows through the term test that $$\lim_{n\to\infty}x_n=0$$

We have $$\log(x)=\int_1^x\frac1t\ dt$$ Thus, $$\begin{align}\int_1^yx\log(x)\ dx&=\int_1^yx\int_1^x\frac1t\ dt\ dx\\\{1<x<y,1<t<x\}&=\int_1^y\int_1^x\frac xt\ dt\ dx\\\{1<t<y,t<x<y\}&=\int_1^y\int_t^y\frac xt\ dx\ dt\\&=\int_1^y\frac{x^2}{2t}\bigg|_{x=t}^{x=y}\ dt\\&=\frac12\int_1^y\frac{y^2}...