Recent content by zaidalyafey

http://advancedintegrals.com/wp-content/uploads/2016/12/advanced-integration-techniques.pdf

Yes as Ssnow suggested. Can you try sovle the integral for n=2 ? Sorry I for the late response I was travelling.

It is a simple differentiation of power $$\frac {d}{dx}[g (x)]^n = n g'(x) [g (x)]^{n-1}$$

Well, I suggest you use integration by parts since $$\frac {\partial }{\partial x} E(-xy)^n =-n\frac {e^{xy}}{x} E(-xy)^{n-1} $$ I think it is easy for small values of n. Try to generalize it.

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For the first equation. Consider the function $$f (x) = \cos(x), a= 0 , b= 2\pi$$ $$2 \pi \sqrt {1 \times 1} >\int^{2\pi}_0\cos(x ) dx=0$$

Consider the function $$f(x) = x(1-x)$$ Then $$\int^1_0 x(1-x) dx = \frac{x^2}{2}-\frac{x^3}{3} = \frac{1}{6} > \frac{f(1)+f(0)}{2} = 0$$

Yah, sure. I forgot to suggest contour integration.

I found a double integral representation $$\gamma= \int^1_0\int^1_0 \frac {x-1}{(1-xy)\log (xy)}dx\, dy$$

To prove that $$\int^\infty_0 t^{s-1}\cos (t) \, dt $$ First note that $$\int^\infty_0e^{-tx} x^{-s} dx= t^{s-1} \Gamma(1-s)$$ We deduce then that $$ \begin {align*} \int^\infty_0 t^{s-1}\cos (t) \, dt&=\frac {1}{\Gamma(1-s)}\int^\infty_0 \int^\infty_0x^{-s}e^{-tx} \cos (t) dx \, dt\\...

$$\lim_{z \to \infty}\mathrm{Cin}(z)-\log z = \gamma$$ Write the integral representation $$\lim_{z \to \infty}\int^z_0 \frac{1-\cos(t)}{t}dt-\log z $$ Can be written $$\lim_{z \to \infty}\int^z_0 \frac{1-\cos(t)}{t}dt-\int^z_0\frac{1}{1+t}dt= \int^\infty_0...

The relation between the psi function and the harmonic numbers becomes handy in solving the Euler sum $$\sum_{k=1}^\infty \frac{H_k}{k^n}$$

You can differentiate z - Ru and integrate ( du / (R² + z² - 2 Rz u)^3/2)

By the way we can prove that $$\psi(s+1) = -\gamma + \int^1_0 \frac {1-x^{s}}{1-x}\, dx$$ Which proves the relation $$\psi (n+1) = -\gamma + H_n $$

You can directly use integration by parts.