Search results
- F( x) = \int\limits _ { 0 } ^ \infty e ^ {-xt/2} ( xt) ^ {- \mu - 1/2 } W _ {\mu + 1/2, \nu } ( xt) f( t) dt,3 KB (490 words) - 13:26, 13 January 2024
- $$I_x(a,b)=\frac1{B(a,b)}\int\limits_0^xt^{a-1}(1-t)^{b-1}dt,$$2 KB (393 words) - 15:50, 31 March 2017
- F(x) = \int_0^\infty \frac{t a(t)}{e^{xt}-1} dt \ .1 KB (182 words) - 21:25, 4 October 2017
- $$F(x)=\int\limits_0^\infty C_\nu(xt)tf(t)dt,$$1 KB (191 words) - 17:37, 1 August 2014
- \[ \sqrt{ ax^2+bx+c} = \pm \sqrt{c}\pm xt.\]2 KB (366 words) - 06:13, 10 April 2023
- \frac{2 e ^ {xt} }{e ^ {t} + 1 }3 KB (477 words) - 08:36, 6 January 2024
- \frac{\phi ( xt) f( t) }{t}3 KB (421 words) - 19:40, 19 January 2024
- $$u_{xt}=0$$5 KB (891 words) - 20:40, 22 December 2018
- where $ {} _ {x} f ( t ) = f ( xt ) $5 KB (797 words) - 09:45, 14 April 2024
- c u _ {xt } = u _ {tt } - {10 KB (1,394 words) - 16:08, 1 April 2020
- $ \Gamma \vdash Xt \Rightarrow \Gamma \vdash Y [ u : = t ] $.11 KB (1,625 words) - 22:11, 5 June 2020
- ...mperature of freezing. This problem has a self-similar solution $ u = u( xt ^ {-} 1/2 ) $,12 KB (1,620 words) - 08:23, 6 June 2020
- where $ M(t)=E(e^{Xt}) $. (This follows10 KB (1,435 words) - 08:56, 25 March 2023
- \mbf{T}^F}{}^x\theta({}^xt^{-1})\mathcal{L}((g,{}^xt),\mathscr{L}^{-1}({}^x\mbf{U})),\\ satisfying $\varphi((g,t) \cdot h) = (g,{}^xt)\cdot \varphi(h)$ for48 KB (8,458 words) - 18:22, 13 August 2023
- \mathop{\rm log} ^ {2} xT + R \ll xT ^ {-1} \36 KB (5,530 words) - 08:29, 14 January 2024
- $$\int_0^\infty\frac{\Xi(t)}{t^2+1/2}\cos xt\,\mathrm{d}t=\frac{\pi}{2}\left[ e^{x/2}-e^{-x/2}\theta(e^{-2x})\right].$$45 KB (7,251 words) - 02:20, 29 June 2022