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F( x) = \int\limits _ { 0 } ^ \infty e ^ {-xt/2} ( xt) ^ {- \mu - 1/2 } W _ {\mu + 1/2, \nu } ( xt) f( t) dt,

$$I_x(a,b)=\frac1{B(a,b)}\int\limits_0^xt^{a-1}(1-t)^{b-1}dt,$$

F(x) = \int_0^\infty \frac{t a(t)}{e^{xt}-1} dt \ .

$$F(x)=\int\limits_0^\infty C_\nu(xt)tf(t)dt,$$

\[ \sqrt{ ax^2+bx+c} = \pm \sqrt{c}\pm xt.\]

\frac{2 e ^ {xt} }{e ^ {t} + 1 }

\frac{\phi ( xt) f( t) }{t}

$$u_{xt}=0$$

where $ {} _ {x} f ( t ) = f ( xt ) $

c u _ {xt } = u _ {tt } - {

$ \Gamma \vdash Xt \Rightarrow \Gamma \vdash Y [ u : = t ] $.

...mperature of freezing. This problem has a self-similar solution $ u = u( xt ^ {-} 1/2 ) $,

where $ M(t)=E(e^{Xt}) $. (This follows

\mbf{T}^F}{}^x\theta({}^xt^{-1})\mathcal{L}((g,{}^xt),\mathscr{L}^{-1}({}^x\mbf{U})),\\ satisfying $\varphi((g,t) \cdot h) = (g,{}^xt)\cdot \varphi(h)$ for

\mathop{\rm log} ^ {2} xT + R \ll xT ^ {-1} \

$$\int_0^\infty\frac{\Xi(t)}{t^2+1/2}\cos xt\,\mathrm{d}t=\frac{\pi}{2}\left[ e^{x/2}-e^{-x/2}\theta(e^{-2x})\right].$$