4
$\begingroup$

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.

$\endgroup$
12
  • $\begingroup$ Is there a condition on "n"? Real, Positive, negative? $\endgroup$
    – rrogers
    Dec 18, 2014 at 14:53
  • $\begingroup$ Good question - I'm only concerned with n > 0 and real. $\endgroup$ Dec 18, 2014 at 15:23
  • $\begingroup$ Well... I pretty much have the answer but haven't accommodated the e^(t(1-s) correctly. Are you in a hurry? $\endgroup$
    – rrogers
    Dec 18, 2014 at 19:37
  • $\begingroup$ Cont: :) Basically you take the Laplace transform divide by p (the target variable) and do the inverse xform. $\endgroup$
    – rrogers
    Dec 18, 2014 at 19:38
  • $\begingroup$ Nope, I'm in no particular hurry. $\endgroup$ Dec 18, 2014 at 19:57

1 Answer 1

2
$\begingroup$

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)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.