# Hypergeometric integral $\int_0^1 dx ~e^{-(\frac{1}{x}+\frac{1}{1-x})}(x)^{b-a-1}(1-x)^{a-1}$.

I have an this integral

$$\int_0^1 dx ~e^{-(\frac{1}{x}+\frac{1}{1-x})}(x)^{b-a-1}(1-x)^{a-1}$$

how can I calculate the integral in form of Bessel function or Hypergeometric function.

As mentioned in Arfken (special function-chapter) $$M(a,c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1 e^{xt}t^{a-1}(1-t)^{c-a-1}dt$$ and $M(a,c;x)= e^x M(c-a,c;-x)$

But I couldn't find useful variable to simplify the integral

• Your integral is indefinite while Bessel function or Hypergeometric function obtain from definite integrals. – Nosrati Nov 26 '17 at 15:04
• Dear @MyGlasses you are right. I have definite integral. I've forgotten write it $\int_0^1 dx$ – asal Nov 26 '17 at 18:35

It seems that the integral can be expressed using McRobert's E-function, which is a generalized hypergeometric function. Let the function defined as $$I(X)=\int_0^X e^{\frac{1}{x}+\frac{1}{X-x}} x^{b-a-1}(X-x)^{a-1}\,dx$$ its value at $X=1$ is to be found. It can be considered as a convolution, its Laplace transform is thus $$\mathcal{L}\left[I;X\to p\right]=F_{b-a-1}(p)F_{a-1}(p)$$ where $$F_\alpha(p)=\int_0^\infty x^\alpha e^{-px-\tfrac{1}{x}}\,dx$$ From Ederlyi TI, p.146 (4.5.29), $$F_\alpha(p)=2p^{-\frac{\alpha+1}{2}}K_{\alpha+1}\left( 2\sqrt{p} \right)$$ $K_\alpha(.)$ is a modified Bessel function. Then $$\mathcal{L}\left[I;X\to p\right]=4p^{-\frac{b}{2}}K_{b-a}\left( 2\sqrt{p} \right)K_{a}\left( 2\sqrt{p} \right)$$ Now, using the inverse Laplace transform of the product of two modified Bessel functions derived in this paper, $$I(1)=\frac{2^{b-2a}}{\sqrt{\pi}}\sum_{i,-i}\frac{1}{i}\left[E\left( \left. \begin{array}{l} b+1,a+1,b-a+1,1\\ \tfrac{b}{2}+1,\tfrac{b+3}{2} \end{array} \right|4e^{i\pi} \right)\right]$$ where the sum is taken above and below the branch cut $\Re z<0$ of the McRobert E-function $E\left( \left. \begin{array}{l} u_p\\ v_q \end{array} \right|z \right)$. This expression may probably be simplified, at least for some specific values of $a$ and $b$ and expressed in terms of classical generalized hypergeometric functions or in terms of Meijer G-function.

Remark 1: in the Ederlyi table (61) p.285, a particular case can be obtained: $$\mathcal{L}^{-1}\left[\sqrt{p}K_{\nu+\tfrac{1}{2}}(2\sqrt{p})K_{\nu-\tfrac{1}{2}}(2\sqrt{p})\right]=\frac{1}{4}\sqrt{\frac{\pi}{2}}e^{-2}W_{\tfrac{1}{2},\nu}(4)$$ where $W_{\mu,\nu}(.)$ is the Whittaker function. Choosing $\nu=-1/2,b=-1,a=-1$, it comes $$I(-1,-1)=\int_0^1 e^{\frac{1}{x}+\frac{1}{1-x}} x^{-1}(1-x)^{-2}\,dx=\sqrt{\frac{\pi}{2}}e^{-2}W_{\tfrac{1}{2},-\tfrac{1}{2}}(4)$$ However, numerically this result is exact if the r.h.s. is multiplied by a factor $\sqrt 2$. Typo in the table or mistake in my derivation?

Remark 2: Changing $x\to 1-x$ in the integral gives the identity $$I(a,b)=I(b-a,b)$$ Moreover, by the relation $(1-x)^a=(1-x)^{a-1}-x(1-x)^{a-1}$, we have the property $$I(a+1,b)=I(a,b-1)-I(a,b)$$ This expression can be generalized, by the use of the binomial expansion of $(1-x)^{a+k}$. With these recurrence relation and the particular cases of @Nemo and from this answer, we can access to many special cases of the integral.

• wow. that sounds great thank you so much dear @Paul Enta – asal Nov 30 '17 at 19:11
• Thanks. I edited to correct a mistake and give a special case of the integral and recurrence relations. – Paul Enta Nov 30 '17 at 20:44

At least there is this particular closed form $$\int_0^1e^{-\left(\frac{1}{x}+\frac{1}{1-x}\right)} \frac{dx}{x^{3/2}(1-x)^{3/2}}=2\frac{\sqrt{\pi}}{e^4}.$$

The general case also should have a closed form and I remember seeing it somewhere...

• This result is a particular case of the relation$$\int_0^1e^{-\left(\frac{1}{x}+\frac{1}{1-x}\right)} x^{\tfrac{b}{2}-1}(1-x)^{\tfrac{b}{2}-1}=2^{-\tfrac{b}{2}}\sqrt{\pi}e^{-2}W_{-\tfrac{b}{4},\tfrac{b}{4}}(4)$$. When $b=-1$, the Whittaker function value is $\sqrt 2 e^{-2}$. This expression is obtained using Laplace inversion formula (59) p. 285 of Ederlyi. – Paul Enta Nov 30 '17 at 21:27