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For \(n\in{\mathbb{N}}\,,\;n>2\,,\) prove that \[\displaystyle \int_{0}^{+\infty}{\frac{\log\bigl({\tfrac{1}{x}}\bigr)}{(1+x)^n}\, dx}=\frac{1}{n-1}\mathop{\sum}\limits_{k=1}^{n-2}{\frac{1}{k}}\,.\]

\[\displaystyle \int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}dx=\int_{0}^{\infty }{x^{\frac{1}{2}-1}e^{-x}dx}=\Gamma\bigl({\tfrac{1}{2}}\bigr)\] From Euler's formula \(\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi\,z)}\) for \(z=\frac{1}{2}\) we have that \begin{align*} \Gamma^2\bigl({\tfrac{1}{2}}\bigr)&...

For \(a>0\): \begin{align*} I&=\int_0^{+\infty} {\frac{\sin^2 x}{x^2 \,(x^2+a^2 )}\, {\rm d}x}\\ &=\frac{1}{a^2}\int_0^{+\infty}{\frac{\sin^2 x}{x^2}\, {\rm d}x}-\frac{1}{a^2}\int_0^{+\infty} {\frac{\sin^2 x}{x^2+a^2}\, {\rm d}x}\\ &=\frac{1}{a^2}\int_0^{+\infty}{\frac{\sin^2 x}{x^2}\, {...

Calculate the integral:

$$\int_{0}^{\infty}{\cos({x^2})\,dx}\,.$$

Calculate the integral:

$$\int_{0}^{\infty}{{\mathrm{e}}^{-{\mathrm{i}}\,x^2}dx}\,.$$

For the values of \(a\) for which the integral converges, calculate the integral \[\displaystyle\int_{0}^{+\infty}{\frac{x^{a}}{1+x^{a}}dx}\,.\]

We give a solution to some point: \begin{align*} I&=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )dx\\ &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\frac{1}{x}} \\ {dx\,=\,-\frac{1}{t^2}\,dt} \end{subarray}}...

Evaluate \[\displaystyle\int_{0}^{1}\log({\Gamma(x+1)})\,dx\,,\] where \(\Gamma(x)\) is the gamma function.

Find \(\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\,.\)

Examine if the series \[\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}{(-1)^{n}\biggl({\frac{1}{\sqrt{n}}+\frac{(-1)^{n-1}}{n}}\biggr)}\] converges conditionally.