User:Proof2016

$$e^{H_n}\ln\left[H_n\left(H_n+\frac{1}{2}\right)\right]-e^{H_n-1}\left(H_n+1\right)< 2n$$; n>=1.

-

$$\int_{0}^{\infty}e^{-x}\ln\left(\frac{1+e^{-x}}{1-e^{-x}}\right)=\ln(4)$$

--

$$\lim_{N \to \infty}N-\int_{0}^{N}e^{xe^{-x}}dx=-\sum_{n=1}^{\infty}\frac{1}{n^{n+1}}$$

- $$k\int_{0}^{\infty}e^{-x}\sin(x^ke^{-x})dx=k\int_{0}^{\infty}e^{-x}\sin(kxe^{-x})dx$$

$$-\int_{0}^{\infty}ae^{-x-ae^{-x}}=e^{-a}-1$$

$$\int_{0}^{\infty}ae^{-x+ae^{-x}}=e^{a}-1$$

$$\int_{0}^{\infty}\left(\frac{x\cos(x)}{e^x-1}-\frac{x\sin(x)\ln(1-e^{-x})}{2}\right)dx=\sum_{n=1}^{\infty}\frac{1}{(n+n^{-1})^2}$$

---

$$-\int_{0}^{\infty}\cos(x)\ln(1-e^{-x})dx=\sum_{n=1}^{\infty}\frac{1}{n^2+1}$$

$$-\int_{0}^{\infty}\cos(x)\ln(1+e^{-x})dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2+1}$$

$$\int_{0}^{\infty}\frac{\sin(x)}{e^x-1}dx=\sum_{n=1}^{\infty}\frac{1}{n^2+1}$$

$$\int_{0}^{\infty}\frac{\sin(x)}{e^x+1}dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2+1}$$

$$-\int_{0}^{\infty}x\sin(x)\ln(1-e^{-x})dx=2\sum_{n=1}^{\infty}\frac{1}{(n^2+1)^2}$$

$$-\int_{0}^{\infty}x\sin(x)\ln(1+e^{-x})dx=2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n^2+1)^2}$$

---

$$\int_{0}^{\infty}xe^{-x}\ln^2(x)dx=\gamma^2-2\gamma+\zeta(2)$$

--

1: $$(x)_1=x$$

$$\int_{0}^{\infty}xe^{-x}\ln(x)dx=1-\gamma$$

2: $$(x)_2=x(x+1)=x+x^2$$

$$\int_{0}^{\infty}x^2e^{-x}\ln(x)dx=1(1)+2(1)-2\gamma$$

3: $$(x)_3=x(x+1)(x+2)=2x+3x^2+x^3$$

$$\int_{0}^{\infty}x^3e^{-x}\ln(xe^{-x})dx=1(2)+2(3)+3(1)-3!\gamma$$

4: $$(x)_4=x(x+1)(x+2)(x+3)=6x+11x^2+6x^3+x^4$$

$$\int_{0}^{\infty}x^4e^{-x}\ln(xe^{-x})dx=1(6)+2(11)+3(6)+4(1)-4!\gamma$$

and son on...

--

$$\int_{0}^{\infty}\sin(xe^{-x})dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2n}}$$

$$\int_{0}^{\infty}\sinh(xe^{-x})dx=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2n}}$$

---

0: $$(x)_0=1$$

$$-\int_{0}^{\infty}x^0e^{-x}\ln(xe^{-x})dx=1+\gamma$$

1: $$(x)_1=x$$

$$-\int_{0}^{\infty}xe^{-x}\ln(xe^{-x})dx=1+\gamma$$

2: $$(x)_2=x(x+1)=x^2+x$$

$$-\int_{0}^{\infty}x^2e^{-x}\ln(xe^{-x})dx=2\gamma+1(1)+2(1)$$

3: $$(x)_3=x(x+1)(x+2)=x^3+3x^2+2x$$

$$-\int_{0}^{\infty}x^3e^{-x}\ln(xe^{-x})dx=3!\gamma+1(1)+2(3)+3(2)$$

4: $$(x)_4=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2+6x$$

$$-\int_{0}^{\infty}x^4e^{-x}\ln(xe^{-x})dx=4!\gamma+1(1)+2(6)+3(11)+4(6)=a_1+2a_2+3a_3+4a_4$$

and son on...

4: $$(x)_4=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2+6x$$

$$-\int_{0}^{\infty}x^4e^{-x}\ln(x^ke^{-x})dx=k(4!\gamma+1(1)+2(6)+3(11)+4(6))-2(4+1)!$$

general:

$$-\int_{0}^{\infty}x^ne^{-x}\ln(x^ke^{-x})dx=k\left(n!\gamma+\sum_{i=1}^{n}ia_i\right)-2(n+1)!$$

Where $$a_i$$ are the coefficients of the $$(x)_n$$

others

$$\int_{0}^{\infty}\frac{2x}{(x^2+1)(x^2+2)^2}dx=-\ln\frac{1}{2}-\frac{1}{2}$$

$$\int_{0}^{\infty}\frac{2x}{(x^2+1)^2(x^2+2)}dx=1-\ln2$$

$$\int_{0}^{\infty}\frac{nx^{n-1}}{(x^n+a)(x^n+b)}dx=\frac{1}{b-a}\ln{\frac{b}{a}}$$

$$\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{ax^{a-1}}{(x^a+n)^k}dx=\frac{\zeta(k-1)}{k-1}$$; k>1

$$\int_{0}^{\infty}\frac{2x}{x^4+2x^2+2}dx=\frac{\pi}{4}$$

$$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$

$$\int_{0}^{\infty}\left(\frac{2x}{x^4+2x^2+2}\right)^2dx=\frac{\pi}{15}$$

-

$$\left(\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{2n+1}\right)^k=(k-1)\sum_{n=-\infty}^{\infty}\frac{(-1)^k}{(2n+1)^k}$$

-

$$\sigma(n)\le e^{H_n}\ln\left(H_n+\frac{1}{2}\right)$$; simplified of Jeffrey-Robin. RH equivalent

-

$$(x)_n=x(x+1)(x+2)...(x+n-1)$$

$$(x)_0=1$$

$$\int_{0}^{\infty}x^{0}e^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\frac{1}{n^n}$$

1: $$(x)_1=x$$

$$\int_{0}^{\infty}xe^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\frac{1}{n^n}$$

2: $$(x)_2=x(x+1)=x^2+x$$

$$\int_{0}^{\infty}x{^2}e^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\left(\frac{1}{n^n}+\frac{1}{n^{n+1}}\right)$$

3: $$(x)_3=x(x+1)(x+2)=x^3+3x^2+2x$$

$$\int_{0}^{\infty}x{^3}e^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\left(\frac{1}{n^n}+\frac{3}{n^{n+1}}+ \frac{2}{n^{n+2}}\right)$$

4: $$(x)_4=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2+6x$$

$$\int_{0}^{\infty}x{^4}e^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\left(\frac{1}{n^n}+\frac{6}{n^{n+1}}+ \frac{11}{n^{n+2}}+\frac{6}{n^{n+3}}\right)$$

and so on... sterling coefficients.

$$\int_{0}^{\infty}x{^k}e^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\frac{(n)_k}{n^{n+k}}$$

-

$$(x)_0=1$$

$$\int_{0}^{\infty}\frac{x^{0}}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{\Gamma(n)}{n^n}$$

1: $$(x)_1=x$$

$$\int_{0}^{\infty}\frac{x}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{\Gamma(n)}{n^n}$$

2: $$(x)_2=x(x+1)=x^2+x$$

$$\int_{0}^{\infty}\frac{x^{2}}{e^x-x}dx=\sum_{n=1}^{\infty}\Gamma(n)\left(\frac{1}{n^n}+\frac{1}{n^{n+1}}\right)$$

3: $$(x)_3=x(x+1)(x+2)=x^3+3x^2+2x$$

$$\int_{0}^{\infty}\frac{x^{3}}{e^x-x}dx=\sum_{n=1}^{\infty}\Gamma(n)\left(\frac{1}{n^n}+\frac{3}{n^{n+1}}+ \frac{2}{n^{n+2}}\right)$$

4: $$(x)_4=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2+6x$$

$$\int_{0}^{\infty}\frac{x^{4}}{e^x-x}dx=\sum_{n=1}^{\infty}\Gamma(n)\left(\frac{1}{n^n}+\frac{6}{n^{n+1}}+ \frac{11}{n^{n+2}}+\frac{6}{n^{n+3}}\right)$$

and so on... sterling coefficients.

$$\int_{0}^{\infty}\frac{x^{k}}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{(n)_k\Gamma(n)}{n^{n+k}}$$

$$\int_{0}^{\infty}x^ke^{-x}\sinh(xe^{-x})dx=\sum_{n=1}^{\infty}\frac{(2n)_k}{(2n)^{2n+k}}$$

$$\int_{0}^{\infty}x^ke^{-x}\cosh(xe^{-x})dx=\sum_{n=1}^{\infty}\frac{(2n-1)_k}{(2n-1)^{2n-1+k}}$$

$$\int_{0}^{\infty}x^ke^{-x}\sin(xe^{-x})dx=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n)_k}{(2n)^{2n+k}}$$

$$\int_{0}^{\infty}x^ke^{-x}\cos(xe^{-x})dx=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n-1)_k}{(2n-1)^{2n-1+k}}$$

general

$$\int_{0}^{\infty}x^ke^{-(a+2)}x\sinh(xe^{-x})dx=\sum_{n=1}^{\infty}\frac{(2n)_k}{(2n+a+1)^{2n+k}}$$

$$\int_{0}^{\infty}x^ke^{-(a+2)x}\cosh(xe^{-x})dx=\sum_{n=1}^{\infty}\frac{(2n-1)_k}{(2n+a)^{2n-1+k}}$$

$$\int_{0}^{\infty}x^ke^{-(a+2)x}\sin(xe^{-x})dx=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n)_k}{(2n+a+1)^{2n+k}}$$

$$\int_{0}^{\infty}x^ke^{-(a+2)x}\cos(xe^{-x})dx=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n-1)_k}{(2n+a)^{2n-1+k}}$$

$$\int_{0}^{\infty}\frac{(1-xe^{-x})e^{-ax-x(b-ce^{-x})}}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{c^{n-1}}{(a+b+n)^n}$$

$$\Gamma(s-1)\int_{0}^{\infty}\frac{1-xe^{-x}}{(e^x-x)^s}dx=\sum_{n=1}^{\infty}\frac{(n+s-3)!}{(n+s-1)^n}$$

$$\Gamma(s)\int_{0}^{\infty}\frac{xe^{-x}}{(e^x-x)^s}dx=\sum_{n=1}^{\infty}(n-1)\frac{\Gamma(n+s-2)}{(n+s-1)^n}$$

$$a\int_{0}^{\infty}\left(\frac{x}{e^{ax}-x}\right)^2dx=\sum_{n=1}^{\infty}\frac{n(n)!}{[a(n+1)]^{n+1}}$$

$$a\int_{0}^{\infty}\left(\frac{x}{be^{ax}-cx}\right)^2dx=\sum_{n=1}^{\infty}\frac{n(n)!c^{n-1}}{[ab(n+1)]^{n+1}}$$

$$\int_{0}^{\infty}\left(\frac{x}{e^{ax}-x^2}\right)^2dx=\sum_{n=1}^{\infty}\frac{n(2n)!}{[a(n+1)]^{2n+1}}$$

$$\int_{0}^{\infty}(x^2-x)e^{-x(1-e^{-x})}dx=\lim_{N \to \infty}N- \int_{0}^{N}\frac{(1-xe^{-x})e^{x(1+e^{-x})}}{e^x-x}dx$$

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$$\int_{0}^{\infty}e^{-x(1-e^{-x})}dx=\sum_{n=1}^{\infty}\frac{1}{n^n}$$

$$\int_{0}^{\infty}e^{-x(1+e^{-x})}dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^n}$$

$$\int_{0}^{\infty}\frac{1}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{(n-1)!}{n^n}$$

$$\int_{0}^{\infty}\frac{1}{e^x+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n^n}$$

-- General

$$\int_{0}^{\infty}\frac{e^{-x}}{x}\left(a+x^se^{-ax}-\frac{1-e^{-ax}}{1-e^{-x}}\right)dx=\ln\Gamma(a+1)+\frac{\Gamma(s)}{(a+1)^s}$$

$$\Gamma(s)\int_{0}^{\infty}\frac{e^{-bx}}{(ae^{cx}-dx)^s}dx=\sum_{n=1}^{\infty} \frac{(d)^{n-1}\Gamma(n+s-1)}{(ac(n+s-1+b))^n}$$

$$\frac{c^{s-2}\Gamma(s)}{b}\int_{0}^{\infty}\frac{1}{(ce^{bx}-ax)^s}dx=\sum_{n=1}^{\infty} \frac{(a)^{n-2}(n+s-2)!}{(bc(n+s-2))^n}-\frac{\Gamma(s-1)}{abc}$$

$$\int_{0}^{\infty}be^{-bx(c-ae^{-bx})}dx=\sum_{n=1}^{\infty}\frac{(a)^{n-1}}{(n+c-1)^n}$$

$$\int_{0}^{\infty}\frac{1}{be^x-ax}dx=\sum_{n=1}^{\infty}\frac{(a)^{n-1}\Gamma(n)}{(bn)^n}$$

sum

$$\sum_{n=1}^{\infty}\frac{e^{-n}}{n}\left(1+n^2e^{-n}-\frac{1-e^{-n}}{1-e^{-n}}\right)= \frac{1}{e^{2}+e^{-2}-2}$$

$$\sum_{n=1}^{\infty}\frac{e^{-n}}{n}\left(1+ne^{-n}-\frac{1-e^{-n}}{1-e^{-n}}\right)= \frac{1}{e^{2}-1}$$

$$\sum_{n=1}^{\infty}x^{n-1}e^{n(1-x)}=\frac{1}{e^x-x-1}$$

other

$$\int_{0}^{\infty}e^{-x(1-e^{-x})}dx=\int_{0}^{\infty}xe^{-x(1-e^{-x})}dx= \sum_{n=1}^{\infty}\frac{1}{n^n}$$

$$\int_{0}^{\infty}\frac{1}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{\Gamma(n)}{n^n}$$

$$\int_{0}^{\infty}\frac{x}{e^x-x}dx=\sum_{n=1}^{\infty}\frac{\Gamma(n)}{n^n}$$

$$\Gamma(s)+\int_{0}^{\infty}\frac{x^{s+1}e^{-x}}{e^x-x}dx=\int_{0}^{\infty}\frac{x^s}{e^x-x}dx$$

-

$$\int_{0}^{\infty}\frac{x^s-1}{x-1}{\cdot}\frac{1-e^{-ax}}{e^{ax}-1}dx=\sum_{n=1}^{s}\frac{n!}{a^n}$$

$$\int_{0}^{\infty}{x^{s-1}}{\cdot}\frac{1-e^{-x}}{e^{x}-1}dx=\Gamma(s)$$

$$\frac{a^s}{\Gamma(s)}\int_{0}^{\infty}{x^{s-1}}{\cdot}\frac{1-e^{-ax}}{1+e^{ax}}dx=2\eta(s)-1$$

---

$$\int_{0}^{\infty}\frac{1-e^{-x}}{x(e^{ax}-1)}dx=\frac{1}{a}H_{\frac{1}{a}}$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}(1-e^{-x})}{x(e^{ax}-1)}dx=\frac{1}{a^s}H_{\frac{1}{a},s}$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}(1-e^{-bx})}{x(e^{ax}-1)}dx=\frac{1}{a^s}H_{\frac{b}{a},s}$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}(1-e^{-bx})}{x(e^{ax}+1)}dx=\frac{1}{a^s}H^{-}_{\frac{b}{a},s}$$

-

$$2\int_{0}^{\infty}\frac{e^{-x}+x-1}{x(e^{x}-e^{-x})}dx=\gamma+\ln{\frac{4}{\pi}}$$

$$4\int_{0}^{\infty}\frac{e^{-x}+x-1}{x(e^{2x}-e^{-2x})}dx=\gamma+\ln\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}$$

$$\int_{0}^{\infty}\frac{e^{-x}-1}{x(e^{2x}+1)}dx=\ln{\frac{4\sqrt{2\pi}}{\Gamma^2\left(\frac{1}{4}\right)}}$$

-

$$\int_{0}^{\infty}\frac{e^{-x}+x-1}{x(e^{ax}-1)}dx=\frac{\gamma}{a}-\ln\left(\Gamma\left(1+\frac{1}{a}\right)\right)$$

$$2\int_{0}^{\infty}\frac{\cosh(x)-1}{x(e^{ax}-1)}dx=\ln\left(\frac{\pi}{a\sin\left(\frac{\pi}{a}\right)}\right)$$

$$2\int_{0}^{\infty}\frac{\cosh(x)-1}{x(e^{ax}+1)}dx=\ln\left(\frac{{4a}\sin^2\left(\frac{\pi}{2a}\right)}{{\pi}\sin\left(\frac{\pi}{a}\right)}\right)$$

-

$$q\int_{0}^{\infty}\frac{e^x-1}{e^{qx}-1}dx=\frac{\pi}{2}\cot\left(\frac{\pi}{q}\right)-\ln(2q)+2\sum_{k=1}^{\lfloor \frac{q-1}{2} \rfloor }\cos\left(\frac{2k\pi}{q}\right)\ln{\sin\left(\frac{k\pi}{q}\right)}$$

$$\int_{0}^{\infty}\frac{e^{2x}-1}{e^{3x}-1}dx=\ln\left(\frac{8}{2+\sqrt{2}}\right)$$

$$\int_{0}^{\infty}\frac{\sinh(x)}{e^{ax}-1}dx=\frac{1}{2}-\frac{\pi}{2a}\cot\left(\frac{\pi}{a}\right)$$

$$2\int_{0}^{\infty}\frac{e^{-x}+x-1}{x(e^{2x}-1)}dx=\gamma-\ln\left(\frac{4}{\pi}\right)$$

$$\int_{0}^{\infty}\frac{e^{-x}+x-1}{x(e^{x}-1)}dx=\gamma$$

$$\int_{0}^{\infty}\frac{e^{-x}+x-1}{x(e^{x}+1)}dx=\ln\left(\frac{4}{\pi}\right)$$

general for odd (2k-1)

$$\int_{0}^{\infty}\frac{e^{-(2k-1)x}+(2k-1)x-1}{x(e^{x}+1)}dx= \ln\left(\frac{2^{2k}}{\pi}\cdot{\frac{(2k-2)!!}{(2k-1)!!}}\right)$$

for even (2k)

$$\int_{0}^{\infty}\frac{e^{-2kx}+2kx-1}{x(e^{x}+1)}dx= \ln{2k \choose k}$$

general for all n

$$\int_{0}^{n}\frac{e^{-nx}+nx-1}{x(e^{x}-1)}dx=n\gamma+\ln\left(\Gamma(n+1)\right)$$

--

$$2\int_{0}^{\infty}\frac{e^{x}-x-1}{x(e^{2x}-1)}dx=\ln(\pi)-\gamma$$

$$\lim_{n \to \infty}H_{n}-\int_{0}^{n}\frac{e^{x}-x-1}{x(e^{x}+1)}dx=\ln(\pi)$$

$$\int_{0}^{\infty}\frac{2xdx}{e^{x^2}+1}=\ln(2)$$

$$\int_{0}^{\infty}\frac{2xdx}{(e^{x^2}+1)^2}=\ln(2)-0.5=\eta(1)-\eta(0)$$

$$\int_{0}^{\infty}\frac{2xdx}{(e^{x^2}+1)-(e^{x^2}+1)^2}=\ln(2)-1=\eta(1)-2\eta(0)$$

--

$$\int_{0}^{\infty}\frac{4xdx}{(e^{x^2}+1)(e^{x^2}+2)}=\ln\left(\frac{4}{3}\right)$$

$$\int_{0}^{\infty}\frac{12xdx}{(e^{x^2}+1)(e^{x^2}+3)}=\ln(2)$$

---

$$\int_{0}^{\infty}\frac{x^ndx}{b^{ax^{n+1}}+b^{n+1}-1}=\frac{1}{a(b^{n+1}-1)}$$

$$\int_{0}^{\infty}\frac{x^ndx}{(2^{ax^{n+1}}+1)(2^{ax^{n+1}}+3)}=\frac{1}{2an(2n+1)}$$

-

$$\int_{0}^{\infty}\sum_{m=1}^{\infty}\frac{dx}{2^{m^s{x}}+1}=\zeta(s)$$

$$\int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{x^{n}dx}{2^{(n+1)^s{x^{n+1}}}+1}=\zeta(s+1)$$

$$\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{dx}{(k^s+1)^x+k^s}=\zeta(s)$$

--

$$\sum_{n=1}^{\infty}\frac{H_{n,s}^s}{n\cdot{2^n}}=\eta(s+1)$$

-

$$H_{n,s}^{-}=\sum_{i=1}^{n}\frac{(-1)^{i-1}}{i^s}$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{(e^x+e^{-x})^2}dx=\frac{\eta(s-1)}{2^s}$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{(-x)^{s-1}}{e^{(n+1)x}+e^{nx}}dx=H_{n,s}^{-}-\eta(s)$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{(-x)^{s-1}}{(e^{(n+1)x}+e^{nx})^2}dx=H_{2n,s}^{-}-\eta(s-1)+(2n+1)\eta(s)+k(n)$$

$$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{(e^x+1)^2}dx=\eta(s)-\eta(s-1)$$

$$\int_{0}^{\infty}\frac{dx}{(e^x+1)^3}=\eta(1)-\eta(0)-\frac{\eta(-1)}{2}$$

$$\int_{0}^{\infty}\frac{dx}{(e^x+e^{-x})^{2n}}=\frac{\Gamma^2(n)}{4\Gamma(2n)}$$

-

Gamma function

$$\int_{0}^{\infty}x^{kn}e^{-x^k}dx=\frac{\Gamma\left(n+\frac{1}{k}\right)}{k}$$

$$\int_{0}^{\infty}x^{kn-1}e^{-x^k}dx=\frac{\Gamma\left(n\right)}{k}$$

$$\int_{0}^{\infty}\sum_{i=1}^{\infty}\frac{x^{{s\cdot}i^s-1}}{e^{x^{i^s}}}dx=\zeta(s)\Gamma(s)$$

--

$$\int_{0}^{\infty}\frac{x}{(e^x+1)^2}dx=\frac{\zeta(2)}{2}-\eta(1)$$

$$\int_{0}^{\infty}\frac{2x^2}{(e^x+1)^2}+\frac{x^2}{(e^x-1)^2}dx=\zeta(3)$$

-

$$\zeta(s+1)\zeta(-s)=-2\sin\left(\frac{s\pi}{2}\right) \int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{x^s}{e^{2nx\pi}-1}dx$$

$$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$

$$\zeta(-2n)=0$$

$$\zeta(2n)=(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$

-

$$\sum_{k=0}^{\infty}\frac{1}{(ak+1)^s}-\sum_{i=1}^{n}\frac{1}{i^s}= \frac{b^s}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{e^{b(an+1)x}-e^{b(an-a+1)x}}dx$$; n=>0

$$\zeta^2(s)=\frac{1}{\Gamma(s)} \int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{x^{s-1}}{e^{nx}-1}dx$$

--

$$\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(n+1)(2n+1)(2n+3){2n\choose n}}=\frac{9\pi^2}{4}$$

$$\sum_{n=1}^{\infty}\frac{3^{n+2}} {n^2{2n \choose n}}=2\pi^2$$

$$\sum_{n=0}^{\infty}\frac{3^{n+3}} {(n+1)(2n+1){2n \choose n}}=(2\pi)^2$$

$$\sum_{n=0}^{\infty}\frac{2^n(n+1)(n^2+2n+2)} {(2n+1)(2n+3){2n \choose n}}=\pi$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+ny+x)}{(n+1)(2n+1)(2n+3){2n\choose n}}=(x-y+1)\cdot{\frac{\pi^2}{8}}+(2x-3y+5)\cdot{\frac{\pi}{2}}-(4x-6y+9)$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^3+n^2+ny+x)}{(n+1)(2n+1)(2n+3){2n\choose n}}=(x-y)\cdot{\frac{\pi^2}{8}}+(2x-3y-3)\cdot{\frac{\pi}{2}}-(4x-6y-5)$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^4+n^3+n^2+ny+x)}{(n+1)(2n+1)(2n+3){2n\choose n}}=(x-y+1)\cdot{\frac{\pi^2}{8}}+(2x-3y+10)\cdot{\frac{\pi}{2}}-(4x-6y+16)$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^5+n^4+n^3+n^2+ny+x)}{(n+1)(2n+1)(2n+3){2n\choose n}}=(x-y)\cdot{\frac{\pi^2}{8}}+(2x-3y-9)\cdot{\frac{\pi}{2}}-(4x-6y-18)$$

--- continued fractions of pi

$$\frac{108}{\pi^4}=1+\frac{1^6}{3+}\frac{2^6}{5+}\frac{3^6}{7+} \frac{4^6}{9+...}$$

$$\frac{108\times{90}}{\pi^8}=1+\frac{1^8}{3+}\frac{2^8}{5+}\frac{3^8}{7+} \frac{4^8}{9+...}$$

--

$$\ln\left(\frac{y}{x}\right)=5k(A-B)$$, k>1

Where

$$A=\sum_{n=1}^{\infty}\frac{y^{-\frac{n}{k}}-y^{-\frac{2n}{k}}-y^{-\frac{3n}{k}}+ y^{-\frac{4n}{k}}}{n(y^{-\frac{5n}{k}}-1)}$$

$$B=\sum_{n=1}^{\infty}\frac{x^{-\frac{n}{k}}-x^{-\frac{2n}{k}}-x^{-\frac{3n}{k}}+ x^{-\frac{4n}{k}}}{n(x^{-\frac{5n}{k}}-1)}$$

--

$$F_{1}=1$$, $$F_{2}=1$$ and $$F_{n+1}=F_{n-1}+F_{n}$$; fibonacci sequence

$$\phi=\frac{1+\sqrt{5}}{2}$$; golden ratio

where $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$

$$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$

$$\prod_{n=1}^{\infty}\left[(-1)^{n+1}\phi{F_{n}}+(-1)^nF_{n+1}\right]^{\frac{1}{n^{s+1}}}=\phi^{-\zeta(s)}$$

$$\prod_{n=1}^{\infty}\left[(-1)^{n+1}\phi{F_{n}}+(-1)^nF_{n+1}\right]^{\frac{(-1)^{n+1}}{n^{s+1}}}=\phi^{-\eta(s)}$$

general

$$\prod_{n=1}^{\infty}\left[(-1)^{n+y+1}{\phi}F_{n^x+y}+(-1)^{n+y}F_{n^x+y+1}\right]^{n^{-x(s+1)}}= \phi^{-\zeta(xs)-y\zeta(x(s+1))}$$

-

$$\pi$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-x)(n^2+n-1)^2} {(x+1)(2n+1)(2n+3){2n \choose n}}=-\left(\frac{x-2}{8}\right)\pi^2+2\pi+9$$

$$\sum_{n=0}^{\infty}\frac{2^nn[4n^3-11n^2+(6x-21)n+10x+8]} {(2n+1)(2n+3){2n \choose n}}=x\pi$$

$$\sum_{n=0}^{\infty}\frac{2^nn[n^2+(6x-1)n+10x-4]} {(2n+1)(2n+3){2n \choose n}}=x\pi$$

$$\sum_{n=0}^{\infty}\frac{2^n(4xn^2+2n-10x+3)} {(2n+1)(2n+3){2n \choose n}}=(2x-1)\cdot\frac{\pi}{2}$$

$$\sum_{n=0}^{\infty}\frac{2^n[3(2x-1)n^2+(10x-3)n+3]} {(2n+1)(2n+3){2n \choose n}}=x\pi$$

$$\sum_{n=0}^{\infty}\frac{2^n(1+n)(c^2-a^2n)} {(2n+1)(2n+3){2n \choose n}}=c^2(\pi-3)+\frac{4-\pi}{2}\cdot{b^2}$$

where

$$c^2=a^2+b^2$$

$$\sum_{n=0}^{\infty}\frac{2^n\left(n-\frac{1}{3}\right)^2(n^2+n-1)} {(2n+1)(2n+3){2n \choose n}}=\pi+\frac{1}{3}$$

$$\sum_{n=0}^{\infty}\frac{2^n\left(n-\frac{2}{3}\right)^2(n^2+n-1)} {(2n+1)(2n+3){2n \choose n}}=\pi-\frac{1}{3}$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-8x-2)(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=x\pi^2+8x-1$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n+8x-2)(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=-x\pi^2+8x+1$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-4)(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=\left(\frac{\pi}{2}-1\right)\left(\frac{\pi}{2}+1\right)$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=1-\frac{\pi^2}{8}$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n+6)(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=(3-\pi)(3+\pi)$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n)(n^2-n-10)(n^2+n-1)^2} {(n+1)(2n+1)(2n+3){2n \choose n}}=-\pi^2+2\pi+9$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n)(n^2-n+6)(n^2+n-1)^2} {(n+1)(2n+1)(2n+3){2n \choose n}}=\pi^2+2\pi+9$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n)(n^2-n-11)(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=\pi$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-12)(n^2+n-1)^2} {(n+1)(2n+1)(2n+3){2n \choose n}}=2\pi-\frac{5\pi^2}{4}+9 $$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-12)(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=\frac{5\pi^2}{4}-9 $$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-12)^2(n^2+n-1)} {(n+1)(2n+1)(2n+3){2n \choose n}}=-10\left(\frac{5\pi^2}{4}-9\right) $$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n+1)^2} {(n+1)(2n+1)(2n+3){2n \choose n}}=\frac{(\pi+8)^2}{8} $$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n-1)^2} {(n+1)(2n+1)(2n+3){2n \choose n}}=\frac{\pi^2}{8} $$

---

$$\sum_{n=0}^{\infty}\frac{2^n(n-0.5)^2(n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=\pi$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2-n-12)(n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=\pi$$

$$\sum_{n=0}^{\infty}\frac{2^nn(n^2-n-12)(n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=2\pi$$

$$\sum_{n=0}^{\infty}\frac{2^nn(n^2+n(\pi^2-2)-10)(n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2+n\pi^2)(n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2$$

$$\sum_{n=0}^{\infty}\frac{{2^n}n^2(\pi^2+n-2)(n^2+n-1)} {(2n+1)(2n+3){2n \choose n}}=10+\pi^0+\pi^1+\pi^2+\pi^3$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2(\pi^3+\pi^2)+n)(n^2+n-1)}{(2n+1)(2n+3) {2n \choose n}}=\pi^0+\pi^1+\pi^2+\pi^3+\pi^4$$

$$\sum_{n=0}^{\infty}\frac{2^n(n^2(\pi^3+1)+\pi^2n)(n^2+n-1)}{(2n+1)(2n+3) {2n \choose n}}=\pi^0+\pi^1+\pi^2+\pi^3+\pi^4$$

Lerch functions

$$\Phi(z,s,a,b,c)=\sum_{n=0}^{\infty}\frac{\left(\frac{z}{c}\right)^n}{(bn+a)^s}= \frac{c}{\Gamma(s)}\int_{0}^{\infty}\frac{t^{s-1}e^{-at}}{c-ze^{-bt}}dt$$

Other integrals

$$\phi=\frac{\sqrt{5}+1}{2}$$

$$\frac{\pi^2}{50}=\int_{0}^{\infty}\frac{x(\phi^3{e^{4x}}-\phi^4{e^{3x}}-\phi^3{e^{2x}}+e^x+2)} {(\phi{e^x})^5-1}dx$$

Harmonics

$$H_{n}=\sum_{i=1}^{n}\frac{1}{i}$$

$$\frac{H_{x-1}}{2}=\sum_{n=1}^{\infty}\left( \frac{1}{(nx)^2-1}+\frac{2}{(nx)^2-2^2}+\frac{3}{(nx)^2-3^2}+...+\frac{x-1}{(nx)^2-(x-1)^2m}\right)$$