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Prove that this is the equation of a straight line;
[tex]-\frac{\zeta(x)\zeta(2)y\Gamma(x)}{\int_0^\infty \frac{u^{x - 1}}{e^u - 1} du \int_0^1 \frac{\ln(1 - x)}{x} dx} - \nabla \cdot (\nabla \times \vec{A}) =[/tex]
[tex]\frac{L^{-1}[\frac{F(t)}{t}] \int_{k = 0}^1 \int_{\theta = 0}^{\pi /2} \frac{d\theta dk}{\sqrt{1 - k^2 \sin^2 \theta}} \prod_{n = 0}^\infty \left(m + \frac{4mx^2}{[\pi(2n + 1)]^2}\right) T_n \cos \theta \sum_{n = 1}^\infty \frac{1}{n^n}}{\int_0^1 \frac{dx}{x^x} \cosh x \frac{\cos n\theta}{2\pi ix} \int_{c - i\infty}^{c + i\infty} e^{st}\int_s^\infty f(u) du ds \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n + 1)^2}}[/tex]
[tex]+ \frac{\oint_C \frac{dz}{z - a} \frac{\pi}{x \sinh \pi x} \sum_{n = 1}^\infty \frac{2^{n/2} \sin(n\pi/4) x^n}{n!}}{|\Gamma(ix)|^2 \frac{2\pi i}{b} e^x \sin x} + 2 + \sum_{i,j} [\sigma_0^m, \sigma_i^{m - 1}] \cdot [\sigma_i^{m - 1}, \sigma_j^{m - 2}] + \int_{\partial S} \omega - \int_S d\omega - \int_0^1 \ln x \ln(1 + x) dx - \eta(2) - 2\eta(1)[/tex],
where ζ is Riemann's zeta function, Γ(x) is the gamma function, ∇ is the del operator, L-1 denotes the inverse Laplace transform, Tn is the nth Chebyshev polynomial of the first kind, C is a simple closed curve bounding a region having z = a as an interior point, σ0m is a simplex of an oriented simplicial complex and [σm, σm - 1] is an incidence number, S is a compact, orientable, differentiable k-dimensional manifold with boundary in En and ω is a (k - 1)-form in En, defined, and C1 at all points of S, and η(x) is Dirichlet's eta function.
[ Credit to Zhylliolom. ]
EASIER CHALLENGE:
[tex]\frac{y(e^{\pi i})}{(acos^2(\int_{}^{}e^{x^2})+asin^2(\int_{}^{}e^{x^2}))b} = \frac{x(z+1)(z-1)}{b(z^2-1)} + \frac{gcd(p, p-1)}{a}[/tex]
Where p is a prime number.
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[tex]-\frac{\zeta(x)\zeta(2)y\Gamma(x)}{\int_0^\infty \frac{u^{x - 1}}{e^u - 1} du \int_0^1 \frac{\ln(1 - x)}{x} dx} - \nabla \cdot (\nabla \times \vec{A}) =[/tex]
[tex]\frac{L^{-1}[\frac{F(t)}{t}] \int_{k = 0}^1 \int_{\theta = 0}^{\pi /2} \frac{d\theta dk}{\sqrt{1 - k^2 \sin^2 \theta}} \prod_{n = 0}^\infty \left(m + \frac{4mx^2}{[\pi(2n + 1)]^2}\right) T_n \cos \theta \sum_{n = 1}^\infty \frac{1}{n^n}}{\int_0^1 \frac{dx}{x^x} \cosh x \frac{\cos n\theta}{2\pi ix} \int_{c - i\infty}^{c + i\infty} e^{st}\int_s^\infty f(u) du ds \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n + 1)^2}}[/tex]
[tex]+ \frac{\oint_C \frac{dz}{z - a} \frac{\pi}{x \sinh \pi x} \sum_{n = 1}^\infty \frac{2^{n/2} \sin(n\pi/4) x^n}{n!}}{|\Gamma(ix)|^2 \frac{2\pi i}{b} e^x \sin x} + 2 + \sum_{i,j} [\sigma_0^m, \sigma_i^{m - 1}] \cdot [\sigma_i^{m - 1}, \sigma_j^{m - 2}] + \int_{\partial S} \omega - \int_S d\omega - \int_0^1 \ln x \ln(1 + x) dx - \eta(2) - 2\eta(1)[/tex],
where ζ is Riemann's zeta function, Γ(x) is the gamma function, ∇ is the del operator, L-1 denotes the inverse Laplace transform, Tn is the nth Chebyshev polynomial of the first kind, C is a simple closed curve bounding a region having z = a as an interior point, σ0m is a simplex of an oriented simplicial complex and [σm, σm - 1] is an incidence number, S is a compact, orientable, differentiable k-dimensional manifold with boundary in En and ω is a (k - 1)-form in En, defined, and C1 at all points of S, and η(x) is Dirichlet's eta function.
[ Credit to Zhylliolom. ]
EASIER CHALLENGE:
[tex]\frac{y(e^{\pi i})}{(acos^2(\int_{}^{}e^{x^2})+asin^2(\int_{}^{}e^{x^2}))b} = \frac{x(z+1)(z-1)}{b(z^2-1)} + \frac{gcd(p, p-1)}{a}[/tex]
Where p is a prime number.
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