Derivative of a generalized hypergeometric function

Let $$f(a)={_2F_3}\left(\begin{array}c1,\ 1\\\tfrac32,\ 1-a,\ 2+a\end{array}\middle|-\pi^2\right).$$ How to find $f'(0)$ in a closed form?

Solution 1:

The function $f(a)$ has the following integral representation in terms of the Struve function $\operatorname{H}_{\nu}{(z)}$:

$$\begin{align} f(a) &={_2F_3}{\left(1,1;\frac32,1-a,2+a;-\pi^2\right)}\\ &=(1+a)\Gamma{(1-a)}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x. \end{align}$$

Proof: The integral representation above is a corollary of the following integral representations of the generalized hypergeometric function ${_2F_3}$ (ref.), the generalized hypergeometric function ${_1F_2}$ (ref.), and the confluent hypergeometric function ${_0F_1}$ (ref.):

$${_2F_3}{\left(a_1,a_2;b_1,b_2,b_3;z\right)}=\frac{\Gamma{\left(b_3\right)}}{\Gamma{\left(a_2\right)}\Gamma{\left(b_3-a_2\right)}}\int_{0}^{1}t^{a_2-1}(1-t)^{-a_2+b_3-1}{_1F_2}{\left(a_1;b_1,b_2;zt\right)}\,\mathrm{d}t;$$

$${_1F_2}{\left(a_1;b_1,b_2;zt\right)}=\frac{\Gamma{\left(b_2\right)}}{\Gamma{\left(a_1\right)}\Gamma{\left(b_2-a_1\right)}}\int_{0}^{1}u^{a_1-1}(1-u)^{-a_1+b_2-1}{_0F_1}{\left(;b_1;ztu\right)}\,\mathrm{d}u;$$

$${_0F_1}{\left(;b_1;ztu\right)}=\frac{2\,\Gamma{\left(b_1\right)}}{\sqrt{\pi}\,\Gamma{\left(b_1-\frac12\right)}}\int_{0}^{1}(1-\xi^2)^{b_1-\frac32}\cosh{\left(2\,\sqrt{ztu}\,\xi\right)}\,\mathrm{d}\xi.$$

Note that the last integral above reduces to an extremely simple trigonometric integral in the special case where $b_1=\frac32$ (which is of course true of the problem at hand).

$$\begin{align} {_2F_3}{\left(1,1;\frac32,1-a,2+a;-\pi^2\right)} &=\frac{\Gamma{\left(2+a\right)}}{\Gamma{\left(1\right)}\Gamma{\left(2+a-1\right)}}\int_{0}^{1}t^{1-1}(1-t)^{-1+2+a-1}{_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)}\,\mathrm{d}t\\ &=\frac{\Gamma{\left(2+a\right)}}{\Gamma{\left(1+a\right)}}\int_{0}^{1}(1-t)^{a}{_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)}\,\mathrm{d}t\\ &=(1+a)\int_{0}^{1}(1-t)^{a}{_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)}\,\mathrm{d}t. \end{align}$$

$$\begin{align} {_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)} &=\frac{\Gamma{\left(1-a\right)}}{\Gamma{\left(1\right)}\Gamma{\left(1-a-1\right)}}\int_{0}^{1}u^{1-1}(1-u)^{-1+(1-a)-1}{_0F_1}{\left(;\frac32;-\pi^2tu\right)}\,\mathrm{d}u\\ &=\frac{\Gamma{\left(1-a\right)}}{\Gamma{\left(-a\right)}}\int_{0}^{1}(1-u)^{-(a+1)}{_0F_1}{\left(;\frac32;-\pi^2tu\right)}\,\mathrm{d}u\\ &=-a\int_{0}^{1}(1-u)^{-(a+1)}{_0F_1}{\left(;\frac32;-\pi^2tu\right)}\,\mathrm{d}u. \end{align}$$

$$\begin{align} {_0F_1}{\left(;\frac32;-\pi^2tu\right)} &=\frac{2\,\Gamma{\left(\frac32\right)}}{\sqrt{\pi}\,\Gamma{\left(\frac32-\frac12\right)}}\int_{0}^{1}(1-\xi^2)^{\frac32-\frac32}\cosh{\left(2\,\sqrt{-\pi^2tu}\,\xi\right)}\,\mathrm{d}\xi\\ &=\frac{2\,\Gamma{\left(\frac32\right)}}{\sqrt{\pi}\,\Gamma{\left(1\right)}}\int_{0}^{1}\cosh{\left(2\,\sqrt{-\pi^2tu}\,\xi\right)}\,\mathrm{d}\xi\\ &=\frac{2\,\Gamma{\left(\frac32\right)}}{\sqrt{\pi}}\int_{0}^{1}\cos{\left(2\,\pi\,\sqrt{tu}\,\xi\right)}\,\mathrm{d}\xi\\ &=\int_{0}^{1}\cos{\left(2\,\pi\,\sqrt{tu}\,\xi\right)}\,\mathrm{d}\xi\\ &=\frac{\sin{\left(2\pi\sqrt{tu}\right)}}{2\pi\sqrt{tu}}. \end{align}$$

Next comes the back-substitution. Insert the result for ${_0F_1}$ into the integral for ${_1F_2}$, and then substitute $u=v^2$:

$$\begin{align} {_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)} &=-a\int_{0}^{1}(1-u)^{-(a+1)}{_0F_1}{\left(;\frac32;-\pi^2tu\right)}\,\mathrm{d}u\\ &=-a\int_{0}^{1}(1-u)^{-(a+1)}\frac{\sin{\left(2\pi\sqrt{tu}\right)}}{2\pi\sqrt{tu}}\,\mathrm{d}u\\ &=-a\int_{0}^{1}(1-v^2)^{-(a+1)}\frac{\sin{\left(2\pi\sqrt{t}\,v\right)}}{\pi\sqrt{t}}\,\mathrm{d}v\\ &=-\frac{a}{\pi\sqrt{t}}\int_{0}^{1}(1-v^2)^{-(a+1)}\sin{\left(2\pi\sqrt{t}\,v\right)}\,\mathrm{d}v. \end{align}$$

The next step after reaching the last line above is where the Struve function finally makes its appearance. The integral in previous line is precisely the form required by this integral representation of the Struve function. Hence,

$$\begin{align} {_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)} &=-\frac{a}{\pi\sqrt{t}}\int_{0}^{1}(1-v^2)^{-(a+1)}\sin{\left(2\pi\sqrt{t}\,v\right)}\,\mathrm{d}v\\ &=-\frac{a}{\pi\sqrt{t}}\frac{\pi^{a+1}}{2}t^{\frac{a}{2}+\frac14}\Gamma{(-a)}\operatorname{H}_{-a-\frac12}{(2\pi\sqrt{t})}\\ &=\frac12\pi^{a}t^{\frac{a}{2}-\frac14}\Gamma{(1-a)}\operatorname{H}_{-a-\frac12}{(2\pi\sqrt{t})}. \end{align}$$

Finally, upon substituting $x=\sqrt{t}$,

$$\begin{align} f(a) &={_2F_3}{\left(1,1;\frac32,1-a,2+a;-\pi^2\right)}\\ &=(1+a)\int_{0}^{1}(1-t)^{a}{_1F_2}{\left(1;\frac32,1-a;-\pi^2t\right)}\,\mathrm{d}t\\ &=(1+a)\int_{0}^{1}(1-t)^{a}\cdot\frac12\pi^{a}t^{\frac{a}{2}-\frac14}\Gamma{(1-a)}\operatorname{H}_{-a-\frac12}{(2\pi\sqrt{t})}\,\mathrm{d}t\\ &=\pi^{a}(1+a)\Gamma{(1-a)}\int_{0}^{1}(1-t)^{a}t^{\frac{a}{2}+\frac14}\operatorname{H}_{-a-\frac12}{(2\pi\sqrt{t})}\,\frac{1}{2t^{1/2}}\mathrm{d}t\\ &=\pi^{a}(1+a)\Gamma{(1-a)}\int_{0}^{1}(1-x^2)^{a}x^{a+\frac12}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x\\ &=\pi^a(1+a)\Gamma{(1-a)}\int_{0}^{1}\left[x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x\\ &=(1+a)\Gamma{(1-a)}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x. \end{align}$$

Using the integral representation,

$$\begin{align} f{(a)} &=(1+a)\Gamma{(1-a)}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x, \end{align}$$

differentiate $f(a)$ with respect to $a$ using the product rule:

$$\begin{align} f^\prime{(a)} &=\frac{d}{da}\left[(1+a)\Gamma{(1-a)}\right]\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x+\\&~~~~~(1+a)\Gamma{(1-a)}\frac{d}{da}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x. \end{align}$$

Evaluating the derivative at $a=0$, and using the fact that $\sqrt{x}\operatorname{H}_{-\frac12}{(2\pi\,x)}=\frac{\sin{(2\pi x)}}{\pi}$, we have:

$$\begin{align} f^\prime{(0)} &=\frac{d}{da}\bigg{|}_{a=0}\left[(1+a)\Gamma{(1-a)}\right]\int_{0}^{1}\sqrt{x}\operatorname{H}_{-\frac12}{(2\pi\,x)}\,\mathrm{d}x+\\&~~~~~\Gamma{(1)}\frac{d}{da}\bigg{|}_{a=0}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x\\ &=\frac{d}{da}\bigg{|}_{a=0}\left[(1+a)\Gamma{(1-a)}\right]\int_{0}^{1}\frac{\sin{(2\pi x)}}{\pi}\,\mathrm{d}x+\\&~~~~~\frac{d}{da}\bigg{|}_{a=0}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x\\ &=\frac{d}{da}\bigg{|}_{a=0}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x. \end{align}$$

Using the Leibniz integral rule,

$$\begin{align} &\frac{d}{da}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x =\int_{0}^{1}\frac{\partial}{\partial a}\left[\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\right]\,\mathrm{d}x\\ &=\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\left[\ln{\left(\pi x(1-x^2)\right)}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}-\frac{\partial}{\partial a}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\right]\,\mathrm{d}x. \end{align}$$

Hence,

$$\begin{align} f^\prime{(0)} &=\int_{0}^{1}\sqrt{x}\left[\ln{\left(\pi x(1-x^2)\right)}\operatorname{H}_{-\frac12}{(2\pi\,x)}-\frac{\partial}{\partial a}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\right]\,\mathrm{d}x\\ &=\int_{0}^{1}\sqrt{x}\left[\ln{\left(\pi x(1-x^2)\right)}\operatorname{H}_{-\frac12}{(2\pi\,x)}-\operatorname{H}_{-\frac12}^{(1,0)}{(2\pi\,x)}\right]\,\mathrm{d}x\\ &=\int_{0}^{1}\left[\ln{\left(\pi x(1-x^2)\right)}\sqrt{x}\operatorname{H}_{-\frac12}{(2\pi\,x)}-\sqrt{x}\operatorname{H}_{-\frac12}^{(1,0)}{(2\pi\,x)}\right]\,\mathrm{d}x\\ &=\int_{0}^{1}\left[\ln{\left(\pi x(1-x^2)\right)}\frac{\sin{(2\pi x)}}{\pi}-\sqrt{x}\operatorname{H}_{-\frac12}^{(1,0)}{(2\pi\,x)}\right]\,\mathrm{d}x\\ &=\frac{1}{\pi}\int_{0}^{1}\ln{\left(\pi x(1-x^2)\right)}\sin{(2\pi x)}\,\mathrm{d}x-\int_{0}^{1}\sqrt{x}\operatorname{H}_{-\frac12}^{(1,0)}{(2\pi\,x)}\,\mathrm{d}x. \end{align}$$

To evaluate the derivative of $f(a)$ with respect to $a$ at $a=0$, it will also be helpful to know that

$$\sqrt{x}\,\operatorname{H}_{-\frac12}^{(1,0)}{(2\pi\,x)}=\frac{\left(2\operatorname{Ci}{(2\pi\,x)}-\operatorname{Ci}{(4\pi\,x)}\right)\sin{(2\pi\,x)}-\left(2\operatorname{Si}{(2\pi\,x)}-\operatorname{Si}{(4\pi\,x)}\right)\cos{(2\pi\,x)}}{\pi}.$$

From here, it is not too difficult to show that the conjectured value proposed by @VladimirReshetnikov is indeed correct:

$$\begin{align} f^\prime(0) &=-\frac{\operatorname{Ci}{(2\pi)}-\operatorname{Ci}{(4\pi)}+\ln{2}}{2\pi^2}-\int_{0}^{1}\sqrt{x}\,\operatorname{H}_{-\frac12}^{(1,0)}{(2\pi\,x)}\,\mathrm{d}x\\ &=-\frac{\operatorname{Ci}{(2\pi)}-\operatorname{Ci}{(4\pi)}+\ln{2}}{2\pi^2}-\\ &~~~~\int_{0}^{1}\frac{\left(2\operatorname{Ci}{(2\pi\,x)}-\operatorname{Ci}{(4\pi\,x)}\right)\sin{(2\pi\,x)}-\left(2\operatorname{Si}{(2\pi\,x)}-\operatorname{Si}{(4\pi\,x)}\right)\cos{(2\pi\,x)}}{\pi}\,\mathrm{d}x\\ &=-\frac{\operatorname{Ci}{(2\pi)}-\operatorname{Ci}{(4\pi)}+\ln{2}}{2\pi^2}-\frac{-3\operatorname{Ci}{(2\pi)}+\operatorname{Ci}{(4\pi)}+2\ln{2\pi}}{2\pi^2}+\frac{\ln{(2)}-2\gamma}{2\pi^2}\\ &=\frac{\operatorname{Ci}{(2\pi)}-\ln{2\pi}-\gamma}{\pi^2}.~~\blacksquare \end{align}$$