Integration over a combination of confluent hypergeometric, power, and exponential functions

I am trying to work out this integral. If there is no closed form, can you think of any approximations to it?

$$\int_0^T e^{a (T-x)} (T-x)^{1+m+n} x^k \, _1F_1\Big(1+n;2+m+n;a (x-T) \Big) \, dx$$

update:

I found a couple of relations that helps. That is the closest I got.

At first step, I can use following relation from [1]:

$$_1F_1\Big(1+n;2+m+n;a (x-T)\Big)=e^{a(x-T)} \, _1F_1\Big(1+m;2+m+n;a(T-x)\Big)$$

so I can write the integral as: $$\int_0^T (T-x)^{1+m+n} x^k \, _1F_1\Big(1+m;2+m+n;a(T-x) \Big)dx$$ with change of variable $y=T-x$, I can rewrite as: $$\int_0^T y^{1+m+n} (T-y)^k \, _1F_1\Big(1+m;2+m+n;ay \Big) \, dy$$ A relation I found on [1] has closed form solution to following integral: $$\int_0^T y^{1+m+n} (T-y)^k \, _1F_1\Big(1+m;2+m+n;\color{red}y \Big)dy=T^{2+m+n+k}\frac{\Gamma(2+m+n)\Gamma(k+1)}{\Gamma(3+m+n+k)} \, _1F_1\Big(1+m;3+m+n+k;T\Big)$$ since $a>1$ would the answer be an upper bound or a good approximation to my integral? How far this would be?

Thanks!

[1] Gradshteyn and Ryzhik, "Table of Integrals, Series, and Products", 2007 (page 821 and 1023)

• In regards to the update: since the function ${}_1F_1(a,b,x)$ is increasing on $x>0$ when $a,b>0$, the closed form you found in Gradshteyn and Ryzhik will be a lower bound for your integral. See this plot, which compares the closed form (in blue) with your integral (in purple) for $1 \leq a \leq 1.5$ with $T=3,k=1,m=1,n=0$. Note the scale on the vertical axis; it is only a good approximation for $a$ very very close to $1$. Apr 11, 2013 at 4:32
• Thanks so much! You helped me a lot :)
– Eln
Apr 13, 2013 at 7:10
• If your willing to work off line I have developed a technique for this type of problem but have not been able to get the correct answers (yet). Dec 30, 2014 at 19:12
• What is the present question? Are you looking for bounds or the integral? By a change of variable it seems to me that you already have the absolute integral. Aug 29, 2015 at 19:31
• I am mainly interested in finding the closed form solution of the integral, if possible to write it.
– Eln
Aug 30, 2015 at 3:02

I can give you an approximation for large $a > 0$. For this I'll assume you mean $\operatorname{Log}(e) = 1$ (an analogous approximation can be found if you mean otherwise).

The asymptotic formula

The largest contribution to the integral comes from a neighborhood of $x=0$, and here we have

$$(T-x)^{1+m+n} \approx T^{1+m+n}$$

and

\begin{align} {}_1F_1\Bigr(1+n,2+m+n,a(x-T)\Bigr) &\approx {}_1F_1(1+n,2+m+n,-aT) \\ &\approx \frac{\Gamma(2+m+n)}{\Gamma(1+m)}\,(aT)^{-1-n}, \end{align}

where the last approximation was given in my answer to this other question of yours.

We therefore have, to first order,

\begin{align} &\int_0^T e^{a (T-x)} (T-x)^{1+m+n} x^k {}_1F_1\Bigl(1+n;2+m+n;a (x-T)\Bigr) \, dx \\ &\qquad \approx \frac{\Gamma(2+m+n)}{\Gamma(1+m)} \,e^{aT} T^m a^{-1-n} \int_0^\infty e^{-ax} x^k\,dx \\ &\qquad = \frac{\Gamma(1+k)\Gamma(2+m+n)}{\Gamma(1+m)} \,e^{aT} T^m a^{-2-n-k} \end{align}

with decreasing relative error as $a \to \infty$. Higher order correction terms can be obtained by using more terms in the binomial series for $(T-x)^{1+m+n}$ and in the asymptotic series for ${}_1F_1$.

Comparison with the numerics

If $I(a)$ is your integral and $A(a)$ is the asymptotic formula I derived, then below is the plot of $$e^{-aT}I(a)$$ in blue and $$e^{-aT} A(a)$$ in purple for $a \in [2,6]$ with $T=3$, $k=1$, $m=1$, $n=0$.

• Thanks Antino for your help! Can you please take a look at the update and let me know what you think? That would be appreciated a lot!
– Eln
Apr 11, 2013 at 3:26

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 $y=a\left(T-x\right)$ , $x=T-\frac{y}{a}=\frac{a\cdot T-y}{a}$, $dx=-\frac{y}{a}$

Adjusting the limits:$0\rightarrow a\cdot T$ , $T\rightarrow0$

Gives

$a^{-(2+m+n+k)}\int^{a\cdot T}_0 y^{1+m+n} (a\cdot T-y)^k \, _1F_1\Big(1+m;2+m+n;y \Big) \, dy$

I have a different copy of G&R and use section 7.613 (Eq: 1)

We have for $\lambda=2+m+n$ , $c-\lambda-1=k$ or $c=k+3+m+n$

$a^{-(2+m+n+k)}\int^{a\cdot T}_0 y^{1+m+n} (a\cdot T-y)^k \, _1F_1\Big(1+m;2+m+n;y \Big) \, dy$

$=a^{-\left(2+m+n+k\right)}\left(a\cdot T\right)^{k+2+m+n}\frac{\Gamma\left(2+m+n\right)\cdot\Gamma\left(k+1\right)}{\Gamma\left(k+3+m+n\right)}\,_{1}F_{1}\left(1+m;k+2+m+n;a\cdot T\right)$

$=T^{k+2+m+n}\frac{\Gamma\left(2+m+n\right)\cdot\Gamma\left(k+1\right)}{\Gamma\left(k+3+m+n\right)}\,_{1}F_{1}\left(1+m;k+3+m+n;a\cdot T\right)$

I don't have a commercial CAS and mine won't simplify the derivative, verifying, without some work; which I will do if you can't.

,