# What is this type of integral substutuion

I am familiar with regular substitution but I am confused on what my book writes here:

Evaluate $\int_{0}^{\infty} x^{m} e^{-ax^{n}} \, dx$, where $m$, $n$, and $a$ are positive constants.

Letting $ax^{n} = y$, the integral becomes

$$\int_{0}^{\infty} \left\{\left(\frac{y}{a}\right)^{1/n}\right\}^{m} e^{-y} \, d\left\{\left(\frac{y}{a}\right)^{1/n}\right\} = \frac{1}{na^{(m + 1)/n}} \int_{0}^{\infty} y^{(m + 1)/n - 1} e^{-y} \, dy = \frac{1}{na^{(m + 1)/n} }\Gamma \left(\frac{m + 1}{n}\right)$$

It is from Schaums outline to advanced calculus I thought I understood regular substitution but this is really confusing me. Like what is going on here? How is this being solved in this way and why keep d( a term) in the integral?

Can anyone please help explain it?

Thanks

• what part do you need to understand? – mint Jan 28 '16 at 2:50
• What do you mean by "regular substitution?" – Mark Viola Jan 28 '16 at 2:50
• By regular I mean, something like f(x)=xsqrt(x^2+10)dx , where we would put w=x^2+10, dw/2=xdx and then write the integral like that, and @mint I want to understand why we don't go to dy and instead write d( x in terms of y) etc – PersonaA Jan 28 '16 at 3:20

## 2 Answers

Is this the bit that is confusing you? $$\int_0^\infty x^m\,e^{-ax^n}\operatorname d x = \int_{0}^{\infty} \left\{\left(\frac{y}{a}\right)^{1/n}\right\}^{m} e^{-y} \operatorname d\left\{\left(\frac{y}{a}\right)^{1/n}\right\}$$

Instead of a separate line saying, "let $y=ax^n$ so $x^m=\left(\frac{y}{a}\right)^{m/n}$ and $\operatorname d x = \frac{y^{(1/n) - 1}}{n\,a^{1/n}}\operatorname d y$" they are just doing it all in situ.

If $y=ax^n$, then $x=(y/a)^{1/n}$. Then, we have

\begin{align} &e^{-ax^n}=e^{-y}\\\\ &x^m=(y/a)^{m/n}\\\\ &dx= \frac1n \frac{y^{1/n-1}}{a^{1/n}}\,dy \end{align}

The integral becomes

$$\int_0^\infty x^m\,e^{-ax^n}\,dx=\int_0^\infty (y/a)^{m/n}\,e^{-y}\,\frac{y^{1/n-1}}{a^{1/n}}\,dy$$

Can you finish now?