Approximating $\int_1^\infty x^a (x-1)^b e^{cx}dx$

Question

After lengthy calculations, I arrived at \begin{align*} \int_1^\infty x^a (x-1)^b e^{cx}dx, \end{align*} which cannot be solved in closed-form. I thus seek to approximate the integral, potentially including special functions. The constants $a,b,c$ may be complex but are chosen such that the integral exists.

Any ideas how to approximate the integral? I thought of rounding the real and imaginary part of $a$ and $b$ to integers. Then, simplifying the first two terms and plenty of partial integrations should give a solution. But that seems to be a very naive idea.

Answer 1

Your integral after substituting $x=z+1$ (and replacing $c$ by $-c$ with $c>0$) becomes $$e^{-c} \int_0^\infty (z+1)^a z^b e^{-cz} \, {\rm d}z = e^{-c}\Gamma(b+1) U(b+1,a+b+2,c) \, .$$ The LHS can be calculated explicitly if $a$ is an integer, since then it becomes - after using the binomial theorem on $(z+1)^a$ - a finite sum $$\frac{e^{-c}}{c^{b+1}} \sum_{n=0}^a \binom{a}{n} \frac{(b+n)!}{c^n} \, .$$

In general however, the second Kummer solution $U$ can be quite complicated and as you are mentioning approximations, have you tried to deploy the saddle-point method i.e. $$\int_0^\infty (z+1)^a z^b e^{-cz} \, {\rm d}z = \int_0^\infty e^{-\left\{cz-a\log(z+1)-b\log(z)\right\}} \, {\rm d}z$$ and then expanding the power $$f(z)=cz-a\log(z+1)-b\log(z) \approx f(z_+) + \frac{f''(z_+)}{2} (z-z_+)^2$$ about its minimum $z_+$? For complex $a,b,c$ with $\Re(c)>0$ the contour $(0,\infty)$ can be deformed to go through that minimum/saddlepoint which is given by one of the solutions $$z_\pm = \frac{a+b-c \pm \sqrt{(a+b-c)^2 + 4bc}}{2c}$$ of $$f'(z)=c-\frac{a}{z+1} - \frac{b}{z} =0 \, .$$ I picked the solution that naturally lies on $(0,\infty)$ for positive parameters. The saddlepoint method then gives the approximation $$\int_0^\infty (z+1)^a z^b e^{-cz} \, {\rm d}z \approx e^{-f(z_+)} \sqrt{\frac{2\pi}{f''(z_+)}} = (z_+ + 1)^a z_+^b e^{-cz_+} \sqrt{ \frac{2\pi}{\frac{a}{(z_+ + 1)^2} + \frac{b}{z_+^2}} }$$ off only about 1-2 percent.

Answer 2

Different elements of solution.

I will consider only real values of $a,b$, with $c$ real and negative for evident convergence concerns.

If $a,b$ are positive integers, there is an easy answer because a function of the form $e^{cx}P_n(x)$ has a primitive function of the same form $e^{cx}Q_n(x)$ with degree$(P_n) \ = \ $ degree$(Q_n)=n$ which is:

$$\int e^{ax}P_n(x)dx = e^{ax}\sum\limits_{k = 0}^n (-1) ^k \frac{1} {a^{k+1}}\frac{d^k}{dx^k}\left(P_n(x)\right) + C\tag{*}$$

Proof: First part of this answer (https://math.stackexchange.com/q/155808) by Robjohn.

See the remark at the bottom for a proof using a less "symbolic" approach.

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In the general case, the different tracks I have obtained using the research tool https://approach0.xyz/ (especially efficient on formulas) were dealing with bounds either different from $1$ or $\infty$. Nevertheless, one of the main indications is the closed form expression of the indefinite integral (= primitive function), as found in the answer by Harry Peter here

$$\int e^{ax}x^b(1-x)^c~dx=\dfrac{x^{b+1}\Phi_1(b+1,-c,b+2;x,ax)}{b+1}+C$$ where $\Phi_1$ can be defined in the following way:

$$\text{For} \ \Re \,c>\Re \,a>0, \ \ \ \Phi _{1}(a,b,c;x,y):=$$ $${\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\int _{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}\,\mathrm {d} t.$$

(see here).

With definite bounds, one finds, from the same Harry Peter here, a formula for

$$\int_0^\infty x^{a-1}(1-x)^{b-1}e^{t-cx}~dx=$$

$$=\dfrac{e^t\Gamma(a)\Gamma(b)M(a,a+b,-c)}{\Gamma(a+b)}-(-1)^be^{t-c}\Gamma(b)U(b,a+b,c)\tag{1}$$

(with an interesting splitting of the integral $\int_0^1+\int_1^{+\infty}$)

For $t=0$, formula (1) is nothing else than the Laplace Transform of $x^a (x-1)^b$, easy to find directly.

It remains to subtract to (1) the value of $\int_1^\infty x^a (x-1)^b e^{cx}dx$ as indicated here.

Remark: A more classical approach to formula (*):

If $I_n:= \int x^n e^{ax}dx$, we have: $$(-1)^na^n\frac{I_n}{n!}=e^{ax} \sum_{k=0}^n(-1)^ka^{k-1}\frac{x^k}{k!}$$

(proofs by Qiaochu Yuan and Olivier Oloa here). Please note that the summation above is a kind of truncated exponential.

Answer 3

Beside the complex functions, we could write $$x^a (x-1)^b =x^a\sum_{n=0}(-1)^{b+n} \binom{b}{n}x^n$$ and, if $\Re(c)<0$, $$\int_1^\infty x^{n+a}\,e^{cx}\,dx=E_{-(a+n)}(-c)$$

Answer 4

$$I(a,b,c)=\int_1^\infty x^a(x-1)^b e^{cx}dx$$ $$I_c=I(a+1,b,c)$$ $$I_{ca}=(a+1)I\tag{1}$$ this is a pde so you could try and solve this, initial conditions with $c=0$ or $b=0$ should make this problem simple. However notice that this integral appear to diverge for $c>0$