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I want to find $$\mu_r' = \int_0^\infty y^r\theta e^{-\theta y}dy $$

A day ago I read something here on MSE about differentiation under the integral sign. I am not sure of how it works, however I tried to differentiate wrt $r$ for $r$ times, but I just get a huge expression. Considering $\theta e^{-\theta y}$ a constant wrt $r$ I get first $ry^{r-1}$ then $y^{r-1}+r(r-1)y^{r-2}$ then $(r-1)y^{r-2}+(r-1)y^{r-2}+(ry^{r-2}+r(r-1)(r-2)y^{r-3})$ and so on.

I can see there must be a pattern, but I can't find it. How can I find the general formula for $\mu_r'$ ?

Edit

I am pretty sure that the solution is something of the form $$\frac{r!}{\theta^r}$$ as this is what you get when doing each case separately

Edit 2

As seen in the answers below my derivative is wrong, still I have no clue on how to solve the problem

2 Answers 2

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You are confusing the derivative with respect to $r$ and $y$ since

$$\frac{d}{dy} y^{r} = r y^{r-1}$$

and

$$\frac{d}{dy^n} y^{r} = r (r-1) \cdots (r-(n-1)) y^{r-n}$$

While the nth derivative with respect to $r$ is

$$\frac{d}{dr^n} y^{r} = \log^n(y) y^{r}$$

To solve the integral

$$ \int_0^\infty y^r\theta e^{-\theta y}dy $$

Now let $\theta y = r$

$$ \frac{1}{\theta^{r}}\int^\infty_0 x^r e^{-x}\,dx $$

Now use integration by parts

$$f(r) = \int^\infty_0 x^r e^{-x}\,dx = r\int^\infty_0 x^{r-1} e^{-x}\,dx$$

This implies $f(r) = rf(r-1)$ which is the definition of the factorial for $r > 0$ , hence $f(r) = r!$ .

This implies

$$ \int_0^\infty y^r\theta e^{-\theta y}dy = \frac{r!}{\theta^r} $$

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    True, that was my mistake! Would you know how to prove the identity that I added in the question? – 2017-01-16
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    @Euler_Salter, see my edit. – 2017-01-16
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    how did you get $$\frac{1}{\theta^r}$$ ? – 2017-01-16
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    @Euler_Salter, let $\theta y = x$ – 2017-01-16
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    perfect I'd say – 2017-01-16
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Note that $\frac{dy^r}{dr}=\log(y)y^r$. Therefore, we have for $r>-1$

$$\frac{d^n}{dr^n} \int_0^\infty y^r e^{-\theta y} \,dy=\int_0^\infty \log^n(y)y^re^{-\theta y} \,dy$$

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    Wait how do we have $ry^{r-1} = \log(y) y^r$ ? – 2017-01-16
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    Oh wait no cause that's not the derivative – 2017-01-16
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    which in my case would be $$\int_0^\infty\left(\log(y)y\right)^r e^{-\theta y}dy$$. Is there a way to solve this? – 2017-01-16
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    It can be written in terms of the Gamma and Polygamma functions. – 2017-01-16