Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having trouble evaluating this indefinite integral; even when I followed it through step by step, the answer I obtain is not correct. I typed it up on this text, it's kind of hard to read (I don't have any experience with LaTeX) but I tried to make it as clear as possible; I'd really appreciate if someone could point out my mistake(s):

$$\int t^3 e^{-t^2}dt$$

Let $u=t^2$; $du=2t~dt$, so $dt=\dfrac{du}{2t}=\dfrac{du}{2u^{1/2}}$. Then

$$\begin{align*} \int t^3 e^{-t^2}dt&=\int u^{3/2} e^{-u} \frac{du}{2u^{1/2}}\\ &=\int u^2 e^{-u}\frac{du}2\\ &=\frac12\int u^2 e^{-u}du\;. \end{align*}$$

Now integrate by parts:

$$\begin{array}{cc} a=u^2&db=e^{-u}du\\ da=2u~du&b=-e^{-u} \end{array}$$

$$\begin{align*} ab-\int b~da&=u^2\left(-e^{-u}\right)-\int\left(-e^{-u}\right)(2u)du\\ &=u^2\left(-e^{-u}\right)+\int e^{-u}(2u)du\\ &=u^2\left(-e^{-u}\right)+2\int e^{-u}u~du\;. \end{align*}$$

Another integration by parts:

$$\begin{array}{cc} a=u&db=e^{-u}du\\ da=du&b=-e^{-u} \end{array}$$

$$\begin{align*} ab-\int b~da&=uu\left(-e^{-u}\right)-\int\left(-e^{-u}\right)du\\ &=u\left(-e^{-u}\right)+\int e^{-u}du\\ &=u\left(-e^{-u}\right)+2\left[u\left(-e^{-u}\right)\right]+C\;. \end{align*}$$

share|improve this question
Try wolfram alpha. Click "show steps". –  Bill Cook Sep 19 '12 at 1:24

1 Answer 1

up vote 2 down vote accepted

You have an algebra error here:

$$\begin{align*} \int t^3 e^{-t^2}dt&=\int u^{3/2} e^{-u} \frac{du}{2u^{1/2}}\\ &=\int u^2 e^{-u}\frac{du}2\;: \end{align*}$$

$\dfrac{u^{3/2}}{u^{1/2}}=u$, not $u^2$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.