Improper integral related to the gaussian one

• Dec 26th 2007, 12:59 PM
Krizalid
Improper integral related to the gaussian one
Evaluate $\int_0^\infty {e^{ - \left( {x^2 + x^{ - 2} } \right)} \,dx}.$

Cheers (Sun)
• Dec 26th 2007, 06:49 PM
mr fantastic
Quote:

Originally Posted by Krizalid
Evaluate $\int_0^\infty {e^{ - \left( {x^2 + x^{ - 2} } \right)} \,dx}.$

Cheers (Sun)

Another easy one? Well, I don't have a lot of time right now and I don't want to be a complete killjoy to others, so I'll outline a solution:

Note that $\displaystyle \int_0^\infty {e^{ - \left( {x^2 + x^{ - 2} } \right)} \,dx} = e^2 \int_0^\infty {e^{ - \left( {x - \frac{1}{x} } \right)^2} \,dx} = e^2 \, I(1)$, where $\displaystyle I(\alpha) = e^2 \int_0^\infty {e^{ - \left( {x - \frac{\alpha}{x} } \right)^2} \,dx}$ (this is of course the old trick).

Note that $\frac{dI(\alpha)}{d \alpha} = 0$ and therefore $I(\alpha) = c$, where c is a constant (this is an interesting result). To get c, use the well known result for $I(0)$.

Therefore: $I(1) = ....$ and so .....
• Dec 26th 2007, 06:55 PM
Krizalid
Quote:

Originally Posted by mr fantastic
Another easy one?

Are you forced to post?

Quote:

Originally Posted by mr fantastic
I don't want to be a complete killjoy to others

Well if someone wants to tackle the problem by his(er) own, probably wouldn't read your post.

--

Differentiating under the integral sign is an option, but I was lookin' for another way.
• Dec 26th 2007, 07:10 PM
mr fantastic
Quote:

Originally Posted by Krizalid
Are you forced to post?

Yet another easy one?

No. (And let's not get into infinite recursion here).

Nor is anyone. Which would make for a pretty boring board.

Quote:

Originally Posted by Krizalid
Well if someone wants to tackle the problem by his(er) own, probably wouldn't read your post.

You're probably right. Mind you, my outline still leaves plenty to do ..... And personally, I always think differentiating under the integral is pretty cool. And a neglected technique in most undergraduate calculus courses ......

--

Quote:

Originally Posted by Krizalid
Differentiating under the integral sign is an option, but I was lookin' for another way.

I just knew you was gonna say that ..... (Rofl) Well, someone elses turn.

Out of interest, why are you posting these integrals? Are they red rags being waved at bulls? My speculation is that at this time of year business is slow and questions like these liven things up a bit (and have good teaching points as well).
• Dec 27th 2007, 12:56 AM
Isomorphism
Quote:

Originally Posted by mr fantastic
Out of interest, why are you posting these integrals? Are they red rags being waved at bulls?

Ditto!

Krizalid, you have a good reputation here. And sometimes you solve the problem yourself immediately after the question. That confuses me :confused:

Do you want someone to solve them OR do you want it to be some kind of a tutorial ? If it's the latter you could probably post it under Calculus Tutorials. Ask TPH or someone else to choose your integrals as the Problem of the Week e.t.c. Or at least specify it while posting.

Now if it is the former, you probably want someone to solve it in a different way(Given the fact that you always know the solution). If so, then I suggest you specify the methods you have tried. This will stop causing confusions.No poster(helpers I mean) would want their time to be wasted,right?

Naturally , I am sure, all of us appreciate a good mathematical discussion. So I request you to be clear so that we can have a healthy "integration" debate and not quarreling in such a nice forum.

Hope you understand what I mean :)

Thank you :D
Iso
• Dec 27th 2007, 06:30 AM
Krizalid
Quote:

Originally Posted by mr fantastic
I always think differentiating under the integral is pretty cool.

Yes, it's nice. For this problem I wanted a solution without it.

Quote:

Originally Posted by mr fantastic
And a neglected technique in most undergraduate calculus courses ......

Yes, probably the teacher wants a "clean" solution.

Quote:

Originally Posted by mr fantastic
Out of interest, why are you posting these integrals?

Just for fun. I already know a solution for it, so I just post it right here if someone else wants to give it a try.

Quote:

Originally Posted by Isomorphism
Do you want someone to solve them OR do you want it to be some kind of a tutorial ?

Quote:

Originally Posted by Isomorphism
you probably want someone to solve it in a different way(Given the fact that you always know the solution). If so, then I suggest you specify the methods you have tried. This will stop causing confusions.No poster(helpers I mean) would want their time to be wasted,right?

Well, if a different approach is posted, that'd be nice.

I'm just posting problems like proposed ones.

--

I hope I've clarified doubts about integrals :D

P.S.: in fact, discussion's title is related to the solution that I'm lookin' for.
• Dec 29th 2007, 11:39 AM
Krizalid
Well, by setting $x=\frac1u$ leads an useful integral to solve the problem.

Now

$\int_0^\infty {e^{ - \left( {x^2 + x^{ - 2} } \right)} \,dx} = \int_0^1 {e^{ - \left( {x^2 + x^{ - 2} } \right)} \,dx} + \int_1^\infty {e^{ - \left( {x^2 + x^{ - 2} } \right)} \,dx} .$

The rest follows.