# Evaluate the improper integral: $\int_o^\infty 1.35 \times 10^{-7} e^{-0.03x}x^4 dx$

Evaluate the improper integral: $$\int_o^\infty 1.35 \times 10^{-7} e^{-0.03x}x^4 dx$$ (Needed for a solid processing example in chemical engineering).

Now according to my textbook, this is simply $4/0.03$.

Anyone have a clue as to how they determined this?

• Are you sure? I additionally used wolfram alpha to check this result for me and it is correct I believe – user2250537 Oct 16 '17 at 20:24
• Hint: Gamma function:$$24=4!=\int_0^\infty e^{-t}t^4~{\rm d}t$$where $t=0.03x$ – Simply Beautiful Art Oct 16 '17 at 20:24

The solution involves the gamma function... • You have a couple of mistakes, but you have a image instead of latex code. – Gabriel Sandoval Oct 17 '17 at 16:26
• Here's the equations from latex. $$u = ax$$ $$x = \frac{u}{a}$$ $$dx = \frac{du}{a}$$ \begin{equation}c\int_0^\infty {e^{ - ax}{x^n}dx} = c\int_0^\infty {{e^{ - u}}{{\left( {\frac{u}{a}} \right)}^n}\frac{{du}}{a}} \end{equation} \begin{equation} c\int_0^\infty {{e^{ - ax}}{x^n}dx} = \frac{c}{{{a^{n + 1}}}}\int_0^\infty {{e^{ - u}}{u^n}du} = \frac{c}{{{a^{n + 1}}}} \underbrace {\int_0^\infty {{e^{ - u}}{u^n}du} }_{n!}\end{equation} $$c\int_0^\infty {{e^{ - ax}}{x^n}dx} = \frac{c}{{{a^{n + 1}}}}n!$$ – Dave Rosenman Oct 17 '17 at 22:24
• The gamma function evaluated in n, is not the factorial of n. $\Gamma (n) = (n-1)!$ – Gabriel Sandoval Oct 17 '17 at 23:09
• You have to use latex code instead of images in your answer in order to let the community to revise you answer if required. – Gabriel Sandoval Oct 17 '17 at 23:18

Don't be afraid. Use integration by parts. $$\int_0^\infty 1.35 \times 10^{-7} e^{-0.03x}x^4 dx= 1.35 \times 10^{-7}\int_0^\infty e^{-0.03x}x^4 dx= \\ 1.35 \times 10^{-7} \\\left( e^{(-0.03 x)} \big(-33.3333 x^4 - 4444.44 x^3 - 444444. x^2 - 2.96296×10^7 x - 9.87654×10^8\big) \Bigg\vert _0^\infty \right)$$

Then, you can use L'Hôpital's rule to compute the limit.

$$\begin{array}\\\displaystyle \int_0^\infty a e^{-bx}x^n dx &=\dfrac{a}{b}\displaystyle\int_0^\infty e^{-y}(y/b)^n dy \qquad y = bx, dx = dy/b\\ &=\dfrac{a}{b^{n+1}}\displaystyle\int_0^\infty e^{-y}y^n dy\\ &=\dfrac{an!}{b^{n+1}}\\ \end{array}$$

For $\displaystyle\int_0^\infty 1.35 \times 10^{-7} e^{-0.03x}x^4 dx$, $a=1.35 \times 10^{-7}, b=0.03, n=4$ so the result is $\dfrac{an!}{b^{n+1}} =\dfrac{4!1.35 \times 10^{-7}}{0.03^5} =133\frac13$ according to Wolfy.