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Let $m=1-c$, $n=2a-c$, $x=\frac{1}{\sqrt{1-z}}$. Then on the left hand side one has $$_2F_1(a,a+\frac{1}{2};c;z)=F\left(\frac{n-m+1}{2},\frac{n-m}{2}+1;1-m;1-\frac{1}{x^2}\right),\tag{1}$$ and on the right hand side: $$2^{-m}x^{n-m+1}(x^2-1)^{m/2}{L}^{m}_{n}(x).\tag{2}$$ Using eq. (15.8.17) one can write eq. $(1)$ as $$... 4 One example can be found on the Wolfram functions site$$F_{2n}=\frac n2 \left(\frac32\right)^{n-1}\int_0^{\pi} \left(1+\frac{\sqrt 5}{3}\cos x\right)^{n-1} \sin x \,dx,$$and another one in this note:$$F_n=\frac1{\sqrt5}\left(\frac{\sqrt5 +1}{2}\right)^n-\frac2\pi \int_0^{\infty}\frac{\sin\frac{x}2}{x}\frac{\cos n x-2\sin x\sin nx}{5\sin^2 x+\cos^2 x}dx.$$... 1 Starting with the limit definition of the Gamma function, we have$$\Gamma(z)=\lim_{n\to \infty}\frac{n!n^z}{\prod_{k=0}^n (z+k)}$$Then, we have$$\begin{align} \log \Gamma (z+1)&=\log z+\log \Gamma(z)\\\\ &=\log z+\lim_{n\to \infty}\left(\log(n!n^z)-\sum_{k=0}^n \log(z+k)\right) \end{align}$$Now, the Digamma function can be written ... 1 Not a closed form, but we can derive a simple recursion formula to ease the calculation of the expression. We notice that the result can be viewed as a polynomial in z with some t-dependent coefficients. If we take$$\left(\frac{1}{\sinh(t)}\frac{d}{dt}\right)^n e^{zt} \equiv \sum_{k=0}^n a_k^{(n)}(t) z^k e^{zt}and apply the differential operator we ... 0 x=\cosh(t) \Rightarrow \frac{1}{\sinh(t)}\frac{d }{dt} = \frac{d}{dx} and x=\cosh(t) \Rightarrow arcosh(x)=t \Rightarrow z\,t =z \,arcosh(x)\Rightarrow z\,t =z \left(\ln(x+\sqrt{x^2-1})\right) . So \begin{align} \left( \frac{1}{\sinh(t)}\frac{d }{dt} \right)^n \left( e^{z t} \right) &= \frac{d^n}{d x^n} \left( e^{z ... 0 HINT Take  z=1 , x= \cosh t \rightarrow \frac{d^n}{d x^n} ({\sqrt{x^2+1}-x} )  ( you can modify for $z \ne 1$)

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$$\left(\frac{1}{\sinh(t)}\frac{\text{d}}{\text{d}t}\right)^n e^{zt} = \left(\frac{1}{\sinh^n(t)}\right)z^n e^{zt}$$ You simply have to derive $n$ times the exponential. Then if you want you can write in the end, using $e^{zt} = \sinh(zt) + \cosh(zt)$ the result as: $$z^n \frac{\sinh(zt) + \cosh(zt)}{\sinh^n(t)}$$ EDIT (because what I wrote above is a ...

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I will consider the case when $n$ is even. The fact that these two solutions are equivalent when $n$ is odd can be proved in an analogous manner. Let $n=2\nu$ be even nonnegative integer, $\lambda=\xi+\frac{1}{2}$ and let's absorb $\sqrt{\beta}$ into the definition of $p$. Then the ratio of the two expressions above can be written as $$\displaystyle\frac{ ... 0 May I expect the closed-form of this integral ? Yes, you may. In fact, the answer is 0, due to the parity of the sine and cosine functions. my Mathematica couldn't make the result even when I tried to put n=2 and n=3. Mathematica has no problem evaluating the integral, even in its hypergeometric form, once the two sine terms have been ... 0 This product of two Gaussian hypergeometric functions can be expressed by a sum over generalized hypergeometric functions {}_{4}F_{3} according to formula 4.3.(14) on page 187 in "Higher Transcendental Functions, Vol. 1" by A. Erdelyi (Ed.). I reproduce it here for your convenience. (I've slightly changed the notation of the original.)$$ ...

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(This is not a comment, but an addendum to hbp's answer.) I'm glad I raised a bounty for this question because, thanks to hbp, we can show that the general quintic can be solved in terms of trigonometric and hyperbolic functions. The general quintic can be reduced to the one-parameter Brioschi form, $$w^5-10cw^3+45c^2w-c^2=0\tag1$$ with solution (see ...

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Thanks to Tito for the nice question. Here is a solution in terms of hyperbolic sine, which may not be what you want. $$\sinh(5t) = 5 \sinh t+ 20 \sinh^3 t + 16 \sinh^5 t.$$ With $x = 2\sinh t$ and $b = -2\sinh 5t$, we have $$x^5+ 5 x^3 + 5 x + b = 0\tag1,$$ which is the $a = 1$ case. So the solution is $$x = 2\sinh t = 2\sinh\left( ... 3 I suspect there's a well defined special function related to this series. Is it so ? Not really. Basically, \text{Exl}(x)=-F'(1), where F(k)=\displaystyle\sum_{n\ge0}\frac{x^n}{n!^k}~.~ The only known values of F are F(0)=\dfrac1{1-x},~F(1)=e^x, and F(2)=I_0\big(2\sqrt x\big). See Bessel function for more information. Neither F, let alone ... 7 Unfortunately, there is no single approach that will lead to robust, accurate, and high-performance implementations across the large universe of special functions. Often, two or more methods must be used for different parts of the input domain, and the necessary research and implementation work may take weeks for elementary functions and months for higher ... 1 With some help from Mathematica:$$G_{2,3}^{3,1}\left(z\left| \begin{smallmatrix} 0,1 \\ 0,0,0 \\ \end{smallmatrix} \right.\right) = -2 z \, {_3F_3}\left(\begin{smallmatrix}1,1,1\\2,2,2\end{smallmatrix}\middle|\,z\right)-\frac{1}{2} \frac{\partial^2}{\partial t^2}\left.{_1F_1}\left(\begin{smallmatrix}t\\1\end{smallmatrix}\middle|\,z\right)\right|_{t=0}+ ...

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I just love little challenges... $\Gamma(t)=\lim \limits_{n \to \infty}\frac{n!n^t}{t(t+1)(t+2)\ldots(t+n)}$ Note the denominator of the fraction is in itself a factorial of sorts. $\lim \limits_{n \to \infty}\frac{n!n^t}{t(t+1)(t+2)\ldots(t+n)}=\lim \limits_{n \to \infty}\frac{n!n^t}{\frac{(t+n)!}{(t-1)!}}=\lim \limits_{n \to ... 0 It can be proved by mathematical induction, it is easy to show that for$m=1$,$\Gamma (1)=1$, Suppose$m=k+1$, then $$\Gamma(k+1)=\lim _{n\rightarrow \infty}\frac{n!n^{k+1}}{\prod _{i=k+1}^{k+1+n}i}=\lim _{n\rightarrow \infty}\frac{n!n^{k}}{\prod _{i=k}^{k+n}i}\cdot \frac{nk}{k+1+n}$$, Let $$a_n=\frac{n!n^{k}}{\prod _{i=k}^{k+n}i}$$ and ... 3 Notice that:$\Gamma(m) = \lim\limits_{n \to \infty} \frac{n! \; n^m}{m \times (m+1) \times \dots \times (m+n)} = \lim\limits_{n \to \infty} \frac{(m-1)! \; n! \; n^m}{(m + n)!} = (m - 1)! \times \lim\limits_{n \to \infty} \frac{n! \; n^m}{(m + n)!}$. Now let's show that$\lim\limits_{n \to \infty} \frac{n! \; n^m}{(m + n)!} = 1$:$\lim\limits_{n \to ...

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We have for $s$ with $\Re{e}(s) > 1$ : $\Gamma(s)\zeta(s)=\int_0^\infty\frac{t^{s-1}}{{\rm e}^t-1}\mathrm dt=\int_0^1\frac{t^{s-1}}{\mathrm e^t-1}\mathrm dt+\int_1^\infty\frac{t^{s-1}}{\mathrm e^t-1} dt$. The second integral is holomorphe in $s$. We take the Taylor series in the first integral. We have for all $t$ with $|t| < 2\pi$ $\frac t{{\rm ... 1 One way to describe what makes translating sine waves special is the following: sine waves$f(t)$have the property that the subspace of the space of functions spanned by their translates$f(t + t_0), t_0 \in \mathbb{R}$is$2$-dimensional, spanned by$\cos t, \sin t$. What other functions have this property? (I'm not going to require periodicity yet; this ... 3$F_{2l}(x)$can be calculated in closed form for$\text{Im}\ x<0using the integral from Gradsteyn and Ryzhik $$\int_0^1u^\lambda P_{2l}(u)du=\frac{(-1)^l\Gamma\left(l-\frac{\lambda}{2}\right)\Gamma\left(\frac{1}{2}+\frac{\lambda}{2}\right)}{2\Gamma\left(-\frac{\lambda}{2}\right)\Gamma\left(l+\frac{3}{2}+\frac{\lambda}{2}\right)},\quad \text{Re}\ ... 3 You can try to Fourier transform your equation, using$$\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{ixy}dx$$your equation becomes:$$(-y^2+\lambda^2)\hat f(y)=1$$So:$$\hat f(y)=\frac{-1}{y²- \lambda²}$$And you need to Fourier transform it back:$$f(x)=-\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{ixy}}{y²- \lambda²}dyTo evaluate this, you can use the ... 1 I found the answer, in fact, we can use the following Lemma. If \Re \lambda >0 , \Re c> 0 and z is in the complex plane cut along z\geq 1, then we have " Bateman's integral" \int_{0}^{1} \frac{u^{c-1}}{\Gamma(c)} \frac{(1-u)^{\lambda-1}}{\Gamma(\lambda)} \, _2F_1(a,b;c; zu) \, du = \frac{1}{\Gamma(c+\lambda)} \, ... 1 One method is the following: \begin{align} I_{x} &= \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b+\nu + \mu)_{n}}{n! \, (c)_{n}} \, x^{n} \, J_{n} \end{align} whereJ_{n} = \int_{0}^{1} y^{n + b + \mu -1} \, (1-y)^{\nu-1} \, dy.Now, \begin{align} J_{n} &= B(n + b + \mu, \nu) = B(b+\mu, \nu) \, \frac{(b+\mu)_{n}}{(b+\mu + \nu)_{n}} \end{align} and ... 2 By considering the contour integral\oint_C d\zeta \frac{\zeta^{-z}}{(1+\zeta)^2} $$about a keyhole contour and using the residue theorem, we may derive the relation$$\left (1-e^{-i 2 \pi z} \right) \int_0^{\infty} dx \frac{x^{-z}}{(1+x)^2} = i 2 \pi \left [\frac{d}{d\zeta} e^{-z \log{\zeta}} \right ]_{\zeta=e^{i \pi}} = i 2 \pi (-z e^{-i \pi} ) e^{-i ... 3 You have to use the definition of Heaviside function to decompose the equation into two ones corresponding tot < t_0$and$t > t_0$respectively: $$\left\{\begin{array}{11} y''(t) + q^2 y(t) = 0,\ t < t_0\\ y''(t) + (q^2 + m^2) y(t) = 0,\ t > t_0\end{array}\right.$$ Each of them is an ordinary harmonic oscillator equation, so one has the ... 0 As was mentioned in the comments, there are several possible generalizations: 1. Dobiński formula: $$B_x=\sum_{k=1}^\infty\frac{k^x}{e\,k!}$$ 2. Cesàro integral representation: $$B_x=\frac{2\,\Gamma(x+1)}{\pi\,e}\int_0^{\pi } \sin (t\,x)\sin \left(e^{\cos t}\sin\sin t\right)e^{e^{\cos t}\cos\sin t}\,dt$$ Interestingly, the results from these two formulae ... 1 As usual please double check. You have essentially solved it except for your change of variable: You have$_1F_1\Big(1+n;2+m+n;a (x-T)\Big)=e^{a(x-T)} \, _1F_1\Big(1+m;2+m+n;a(T-x)\Big)\int_0^T (T-x)^{1+m+n} x^k \, _1F_1\Big(1+m;2+m+n;a(T-x) \Big)dx$Now we apply$y=a\left(T-x\right)$,$x=T-\frac{y}{a}=\frac{a\cdot T-y}{a}$,$dx=-\frac{y}{a}... 2 Consider instead \begin{align}\int_0^{\infty}e^{-x}G(x,t)G(x,s)dx&=\frac{1}{(1-t)(1-s)}\int_0^{\infty}e^{-\left(1+\frac{t}{1-t}+\frac{s}{1-s}\right)x}dx=\\ &=\frac{1}{(1-t)(1-s)}\int_0^{\infty}e^{-\frac{1-st}{(1-s)(1-t)}x}dx=\\ &=\frac{1}{1-st}. \end{align} What is the coefficient ofs^mt^n$in the expansion of the right side when$m\neq n$? ... 2 For this post, I'll assume all parameters are positive. Given this, we can substitute$t = k_1 \rho$to get an integral of the form $$I_m(a, c) = \int_0^c J_m(at)\,J_{m+1}(t) \; dt.$$ Now recently my roommate and I worked on the case$m = 0$, so from here on out I'll assume$m = 0$. However, everything I say below works for any$mexcept for one section ... 5 This is only a partial solution. I haven't obtained any closed form, but reduced to an integral of an elementary function. One can simplify the above integral as follows $$I=\int_0^1 \frac{\displaystyle \text{Ai}\left(-\frac{t}{\sqrt[3]{12(1 -t)}}\right)^2+\text{Bi}\left(-\frac{t}{ \sqrt[3]{12(1- t)}}\right)^2}{ \sqrt[6]{1- t}} \, dt.$$ Using the integral ... 4 Integrating by parts we get $$\mathcal{I}_n = \int_{-\infty}^\infty \frac{\operatorname{Ai}^2(x+a_n)}{x^2}dx =2\cdot \int_{-\infty}^\infty \frac{\operatorname{Ai}(x+a_n)\operatorname{Ai}^\prime(x+a_n)}{x}dx\qquad (1)$$ [due to the asymptotics $$\operatorname{Ai}(x)\sim \frac{e^{-\frac{2}{3}x^{3/2}}}{2\sqrt{\pi}x^{1/4}},\qquad \operatorname{Ai}(-x)\sim ... 2 I did some googling, and found that a good (slightly old) reference for this kind of questions is G.N. Watson's “A treatise on the theory of Bessel functions”. In the 1922 edition that I currently have access to, the relevant theory appears in Chapter XV. See particularly 15.22, 15.4 and 15.81. This does not answer all the questions, of course, ... 1 It converges by contour integration (quarter cake slice in the lower right quadrant, no residues): \int_0^\infty dz\, z^{n-1} e^{-iz} + \int_{-i \infty}^0 dz\, z^{n-1} e^{-iz} + + \lim_{a \rightarrow 0^-}\lim_{R \rightarrow \infty}\int_a^{-\pi/2}d\phi \,i R e^{i \phi} R^{n-1} e^{i(n-1) \phi} e^{-i R \cos \phi} e^{R \sin \phi} = 0 so \int_0^\infty ... 0 Using the substitution s(t) = \cosh(t) and write y(t) = Y(s(t)). We thus find for Y the differential equation $$(1-s^2)Y''(s) - s Y'(s) + \left(\frac{2}{1-s^2}+\lambda^2\right) Y(s) = 0,$$ which is (a form of the) the associated Legendre equation. In this case, we can obtain the ‘proper’ associated Legendre equation by ... 2 The alternative definition (easily proven equivalent to the one you give for \lvert z \rvert < 1)$$ \operatorname{Li}_2 (z) = \int_z^0 \frac{\log{(1-t)}}{t} \, dt, $$where the integral can be taken to be along the ray joining 0 to z for z \notin (1,\infty) , shows that the dilogarithm has a branch point at z=1 (since the logarithm inside the ... 1 Adding to the previous answers, let y=f(x)=e^x+x,\;x\in\mathbb{R}^+.$$\begin{align*} f(x)&=y \\ e^x+x&=y \\ e^{e^x+x}&=e^y \\ e^xe^{e^x}&=e^y \\ e^x&=W(e^y) \\ x&=\log W(e^y) \\ f^{-1}(x)&=\log W(e^x)\end{align*}$$One property of the Lambert W-function is that \log W(z) =\log z-W(z) for positive z in the principal real ... 6 To supplement the answer posted by @HenningMakholm, I thought it would be instructive to derive the inverse function in terms of Lambert's W Function. To that end, we proceed. First, we note that$$y=x+e^x\implies e^x=y-x \tag 1$$Then, multiplying both sides of (1) by e^{y-x} reveals$$e^y=(y-x)e^{y-x} \tag 2$$Using the definition of ... 7 That inverse function cannot be expressed by the usual elementary functions, but Wolfram Alpha finds it to be$$ f^{-1}(y) = y - W(e^y) $$where W is Lambert's W function. This at least gives it a name, but may not otherwise seem like much progress because the W function is itself defined as the inverse of xe^x. So in some sense all we have achieved is ... 0 The answer is negative.$$ \int_{0}^{+\infty}\frac{t^{s-1}e^{-t}}{e^{t}-1}\,dt = \Gamma(s)\left(\zeta(s)-1\right) $$if \text{Re}(s)>1. The integrand function has a non-integrable singularity in a right neighbourhood of the origin if \text{Re}(s)\leq 1, hence it doesn't make sense to ask for the value of the LHS for \text{Re}(s)=\frac{1}{2}. 2 Because OP has never said f(t) has to be a continuous function, there are more possible functions than a constant zero. For example, if we define$$f(t)=1\text{ fort\in\Bbb Z$}, 0\text{ otherwise}$$We get a nonconstant function integral of which is 0 on every integral. More generally, any function which is non-zero on a discrete set of points will ... 3 Let's denote the continuous function:$$G(x)=\int_0^x f(t) dt$$Then by the Fundamental Theorem of Calculus,$$\frac{d}{dx}G(x)=f(x)$$When G(x)=C,$$\frac{d}{dx}G(x)=0=f(x)$$1 Clearly the constant is zero,Because:$$ constant = \int_0^0 f(x) dx = 0$$Can you conclude now? 7 Differentiating both sides of$$\int_0^x f(t) dt=constant$$using Fundamental Theorem Of Calculus, as mentioned by Omnomnomnom in his comment,( or Leibnitz Rule as I had earlier written), we get$$f(x)=0$\$

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