I assume you've tried integrating by parts, and found that if you start with the x^2n, it just gets worse, and you can't start with the e^{-ax^2}, because it doesn't have a nice anti-derivative?
Hint: can you split the product in some other way, so that you can get an anti-derivative involving the e^{-ax^2}?
well, i would say sth like this would put u on the right track
[tex]\int_{-\infty}^{\infty}x^{2n}*e^{-ax^2} = \int_{-\infty}^{\infty}x^{2n-1}xe^{-ax^2}dx [/tex] then performing integration by parts by letting
[tex]u=x^{2n-1},v=\int xe^{-ax^2}dx[/tex]
I think this would work. By performing integ by parts a couple of times, i think you will be able to notice some pattern, then use induction to prove the general case.
P.S. This is basically what tiny-tim's Hint says...i hope at least...lol...
The trick is to recognise the difference between a problem and a solution.
In this case, the problem is that you look at ∫e^-ax^2
and think
"I can't integrate that … if only it had an extra 2ax in front of it!"
and change that round to
"I can integrate that if i put an extra 2ax in front of it!"
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling We Value Civility
• Positive and compassionate attitudes
• Patience while debating We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving