# Math Help - integral (x^6 * e^(-x^7)) from -infinity to infinity

1. ## integral (x^6 * e^(-x^7)) from -infinity to infinity

Can someone help me solve this integral?: http://www.wolframalpha.com/input/?i=integral+(x^6+*+e^(-x^7))+from+-infinity+to+infinity

I get a final answer of 0 but the correct answer is positive infinity. I would just like to know how to get there and what I am doing wrong. (My work is attached)

Any help would be greatly appreciated!

2. So we have:

$\int_{-\infty}^{\infty} x^{6}e^{-x^{7}}dx$

Using a simple u-substitution, we get

$\int_{-\infty}^{\infty} e^{u}du$

We have a case where both of our bounds will lead to an improper integral, so its best to split this up (for your own sanity, although you could just as easily keep it the same since its not discontinuous over the reals). It's arbitrary where we split the integral, but - as you see - its best to choose 0 since it'll tide up nicely so:

$\lim_{t \to-\infty}-\frac{1}{7}e^{-x^{7}}|_t^0 + \lim_{t \to \infty}-\frac{1}{7}e^{-x^{7}}|_0^t$

$-1 + \lim_{t \to-\infty}\frac{1}{7}e^{-t^{7}} + \lim_{t \to \infty}\frac{1}{7}e^{-t^{7}} - 1$

Remember the graph of the exponential function, as it approaches negative negative values of "x" it approaches zero, and as it apporaches positive values of "x" it approaches infinity. In one of our limits, we have "t" approaching negative infinity: a negative times a negative is a positive; thus the exponential is being raised to "positive infinity". In the other, our exponential is being raised to a negative, thus, as it gets larger negative, the values approach zero. And so we have:

$-1 + \infty + 0 - 1 \Rightarrow \infty$

3. In looking at your work, it looks like your mistake was made on the second line (I'm looking at the last attempt in your notes), somehow you've lost that your "dummy variable" is approaching two different limits. When you simplified, you combined them together, presuming that the exponentials were able to cancel out. They don't.