How to calculate $\int(5x)^4e^{-2x}dx$ without using IBP?

Question

How do I integrate this?

(UPDATE: Sorry, I should have clarified that I've been told to use the gamma function and not just integration by parts, also that this is a definite integral.)

$$\int_{x=0}^\infty (5x)^4e^{-2x}dx$$

I know that that $\displaystyle\int_0^\infty w^se^{-w}dw = s!=\Gamma(s+1)$, and I've been told to apply this fact, but I can't figure out how to apply this second equation to the first equation above.

I can simplify the first equation as follows:

$$= 5^4\int_{x=0}^\infty x^4e^{-2x}\,dw $$

Then if I let $w = 2x$, I can get to:

$$= \frac{5^4}{2^4} \int_{x=0}^\infty w^4 e^{-w} \, dw$$

But I'm not sure what to do next!

Answer 1

You are getting there just continue your work using IBP not via substitution. But to simplify your computation note that the integral after putting the constant outside the integral sign satisfies a special type of IBP which is via tabular method. Just comment if you have questions.

I got $-e^{-2x}(\frac{625}{2}x^4+625x^3+\frac{1875}{2}x^2+\frac{1875}{2}x+\frac{625}{4})+C$.

I think you must have a definite integral if you want to use the given formula.

Answer 2

Hint: Evaluate $I(a)=\displaystyle\int e^{ax}~dx$, and then repeatedly differentiate both sides with regard to a.

Answer 3

I think you forgot to change the limits also. When one changes the variable of integration he has to change the limits accordingly. The new limits will be the range of the function $f(x)=2x$ in the domain $(0,\infty)$ which is $(0,\infty)$, thus your last expression should be $$= \frac{5^4}{2^4} \int_{w=0}^\infty w^4 e^{-w} \, dw.$$

Can you conclude from here?

Answer 4

You are so close! Notice that since $w=2x$, $$ \frac{5^4}{2^4}\int_{x=0}^\infty w^4 e^{-w}{dw} = \frac{5^4}{2^4}\int_{{\color{green}w}=0}^\infty w^4 e^{-w}{dw} $$ And now you substitute your known fact $$ \frac{5^4}{2^4}\int_{w=0}^\infty w^4 e^{-w}{dw} = \frac{5^4}{2^4}4! $$