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Good morning Grigoris. First I will rewrite the given integral as: \( \displaystyle -\int_{0}^{\infty }\frac{\ln x}{\left ( 1+x \right )^n}\, {\rm d}x \). Now let me consider the integral \( \displaystyle \int_{0}^{\infty }\frac{x^{m-1}}{\left ( 1+x \right )^n}\, {\rm d}x \) whereas \( m\neq n \). T...

Perhaps a well known integral, but anyway.

Evaluate the integral:

$$\int_{0}^{\pi}x\ln \left ( \sin x \right )\, {\rm d}x$$

Evaluate the following integral:

$$\int_{0}^{\infty }\frac{1}{\sqrt{x}}e^{-x}\, {\rm d}x$$

We have that: $$\int_{0}^{\infty}\cos x^2 \, {\rm d}x = \mathfrak{Re}\left ( \int_{0}^{\infty}e^{-ix^2}\, {\rm d}x \right )= \frac{1}{2}\sqrt{\frac{\pi}{2}}$$ The latter integral has been proved here . Note, also, that: $$\int_{0}^{\infty}\sin x^2 \, {\rm d}x = \mathfrak{Im}\left ( \int_{0}^{\infty}...

We will use contour integration on a wedge shaped contour of angle $\widehat{\omega} =\pi/4$ as our contour and an infinite large radius $R$. We are integrating on this contour the function $f(z)=e^{-iz^2}$. The contour is shown at the image below: wedge shaped contour.jpg [/centre] Clearly $f$ has ...

Good evening Grigoris. We apply the sub $x=y^{1/a}$ , thus $dx=\frac{1}{a}y^{1/a-1}dy$. So, if $a<0: x=0\Rightarrow y=\infty , \quad x=\infty \Rightarrow x=0$ , else if $a>0: x=0\Rightarrow y=0, \quad x=\infty \Rightarrow y=\infty$. Therefore: $$\begin{align*} \int_{0}^{\infty} \frac{x^a}{1+x^a}\, {...

Let \(\displaystyle I=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )\, {\rm d}x\). If \(\displaystyle e^{I+1}=\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n\) where \(\displaystyle a, b\) are coprime numbers, then calculate the sum...

Let \(a_1, a_2, \; \dots, a_9 \) be real numbers. Evaluate the \( 10 \times 10 \) determinant: $$\begin{vmatrix} -1 & 1 &1 &1 &1 &1 &1 &1 &1 &a_1 \\ 1& -1 & 1 & 1& 1 & 1& 1 & 1 &1 &a_2 \\ 1&1 & -1 &1 &1 &1 &a...

Let \( V_n(R) \) be the volume of the sphere of center \( 0 \) and radius \( R \) in \( \mathbb{R}^n \). Prove that for \( n \geq 3 \) holds:

$$V_n(1)= \frac{2\pi}{n} V_{n-2} (1)$$

[Hint: Apply Cavaliere's principal twice and after you prove \( V_n (R)= R^n V_n(1) \) convert to polar coordinates ]

Let $a>0$.Evaluate the integral:

$$\int_0^\infty \frac{\sin^2 x}{x^2 \left(x^2+a^2 \right)}\, {\rm d}x$$