Integral of a Recursive Function

[CivilSigma]

Homework Statement

Find the following:

$$ \int_0^\inf \frac{x^{a+1}e^{-x/\delta}}{\delta^{a+1}\Gamma(a+1)} dx; \, a > 1 , \delta >0 , 0 \leq x \leq \inf$$

Homework Equations

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The Attempt at a Solution

The numerator in the integral is constant, so it can be taken outside the integral. I then attempted to solve the numerator through integration by parts.

Letting
$$u=x ; du =dx$$
$$ dv = e^{-x/ \delta}dx ; v = -\delta e^{-x/ \delta} $$

I get the following:

$$\frac{1}{\delta^{a+1}\Gamma(a+1)} \cdot ( -\delta x^{a+1}e^{-x\delta} + \delta \int_0^\inf (a+1)x^{a} e^{-x/\delta}dx)$$

I can see that the integral will keep on iterating for ever, so any ideas on how I can solve this?

Thank you !

 

[fresh_42]

I think the simple substitution $u:=\frac{x}{\delta}$ does the job. You will get a constant factor and $\Gamma(a+2)$ in the numerator.

 

[Ray Vickson]

Homework Statement

Find the following:

$$ \int_0^\inf \frac{x^{a+1}e^{-x/\delta}}{\delta^{a+1}\Gamma(a+1)} dx; \, a > 1 , \delta >0 , 0 \leq x \leq \inf$$

Homework Equations

-

The Attempt at a Solution

The numerator in the integral is constant, so it can be taken outside the integral. I then attempted to solve the numerator through integration by parts.

Letting
$$u=x ; du =dx$$
$$ dv = e^{-x/ \delta}dx ; v = -\delta e^{-x/ \delta} $$

I get the following:

$$\frac{1}{\delta^{a+1}\Gamma(a+1)} \cdot ( -\delta x^{a+1}e^{-x\delta} + \delta \int_0^\inf (a+1)x^{a} e^{-x/\delta}dx)$$

I can see that the integral will keep on iterating for ever, so any ideas on how I can solve this?

Thank you !

Change variables until you can put your integral in the form $\int_0^\infty y^{a+1} e^{-y} \, dy.$ That is an integral that can be evaluated algebraically if $a \geq 0$ is an integer; for other values of $a$ it is not an elementary function, but is a well-known, standard mathematical function nevertheless.

BTW: if you want the $\infty$ sign in LaTeX, just write "\infty"; writing "inf" (without the backslash) won't work, and inserting a backslash to get "\inf" produces $\inf$, which is the mathematical shorthand symbol for the word "infimum".

 

[CivilSigma]

Thank you guys ! I solved the integral after making the substitution y =x/delta and integrating by parts twice.

The final answer I got is:

$$ \delta \cdot (\alpha+1)$$

I never knew about the Gamma function! It's very convenient to say the least :)