# Any way to simplify integral of Confluent Hypergeometric Function of the First Kind?

The integral is this:

$$\int_{-\log n}^{0}e^{t(1-s)} \cdot z \cdot {}_1F_1(1-z, 2, t) dt$$

Is there a way to write this in terms of special functions that eliminates the integral and doesn't use infinite sums either?

It can also be written as

$$-1+\sum_{k=0}^\infty \binom{z}{k}(1-s)^{-k} P(k, (s-1)\log n)$$

where $P(a,z)$ is the regularized lower incomplete gamma function, if that is useful. But I'd really like an expression that doesn't involve integrals or sums.

• Is there a condition on "n"? Real, Positive, negative? Dec 18, 2014 at 14:53
• Good question - I'm only concerned with n > 0 and real. Dec 18, 2014 at 15:23
• Well... I pretty much have the answer but haven't accommodated the e^(t(1-s) correctly. Are you in a hurry? Dec 18, 2014 at 19:37
• Cont: :) Basically you take the Laplace transform divide by p (the target variable) and do the inverse xform. Dec 18, 2014 at 19:38
• Nope, I'm in no particular hurry. Dec 18, 2014 at 19:57

Case $1$: $s=1$

Then $\int_{-\log n}^0z~_1F_1(1-z,2,t)~dt$

$=[-~_1F_1(-z,1,t)]_{-\log n}^0$ (according to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/21/01/01)

$=~_1F_1(-z,1,-\log n)-~_1F_1(-z,1,0)$

$=~_1F_1(-z,1,-\log n)-1$

Case $2$: $s=2$

Then $\int_{-\log n}^0e^{-t}z~_1F_1(1-z,2,t)~dt$

$=\int_{-\log n}^0z~_1F_1(z+1,2,-t)~dt$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Kummer.27s_transformation)

$=[-~_1F_1(z,1,-t)]_{-\log n}^0$ (according to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/21/01/01)

$=~_1F_1(z,1,\log n)-~_1F_1(z,1,0)$

$=~_1F_1(z,1,\log n)-1$

Case $3$: $s\neq1,2$

Then $\int_{-\log n}^0e^{t(1-s)}z~_1F_1(1-z,2,t)~dt$

$=\int_{-\log n}^0\sum\limits_{m=0}^\infty\dfrac{z(1-z)_mt^me^{t(1-s)}}{(2)_mm!}dt$

$=\left[\sum\limits_{m=0}^\infty\sum\limits_{k=0}^m\dfrac{(-z)_{m+1}t^ke^{t(1-s)}}{(m+1)!k!(s-1)^{m-k+1}}\right]_{-\log n}^0$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\sum\limits_{m=0}^\infty\dfrac{(-z)_{m+1}}{(m+1)!(s-1)^{m+1}}-\sum\limits_{m=0}^\infty\sum\limits_{k=0}^m\dfrac{(-z)_{m+1}(-1)^k(\log n)^kn^{s-1}}{(m+1)!k!(s-1)^{m-k+1}}$

$=\sum\limits_{m=1}^\infty\dfrac{(-z)_m}{m!(s-1)^m}-\sum\limits_{k=0}^\infty\sum\limits_{m=k}^\infty\dfrac{(-z)_{m+1}(-1)^k(\log n)^kn^{s-1}}{(m+1)!k!(s-1)^{m-k+1}}$

$=\sum\limits_{m=0}^\infty\dfrac{(-z)_m}{m!(s-1)^m}-1-\sum\limits_{k=0}^\infty\sum\limits_{m=1}^\infty\dfrac{(-z)_{m+k}(-1)^k(\log n)^kn^{s-1}}{(m+k)!k!(s-1)^m}$

$=\left(1-\dfrac{1}{s-1}\right)^z-1+\sum\limits_{k=0}^\infty\dfrac{(-z)_k(-1)^k(\log n)^kn^{s-1}}{(k!)^2}-\sum\limits_{k=0}^\infty\sum\limits_{m=0}^\infty\dfrac{(-z)_{m+k}(-1)^k(\log n)^kn^{s-1}}{(m+k)!k!(s-1)^m}$ (according to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F0/02)

$=\dfrac{(s-2)^z}{(s-1)^z}-1+n^{s-1}~_1F_1(-z,1,-\log n)-n^{s-1}\Phi_1\left(-z,1,1;\dfrac{1}{s-1},-\log n\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)