The hard integral thread.

You rely on the fact the under certain conditions (see end of post), the following equality holds:



You can then do something like this for integrals that are easy:

1. Easy integrals:



so that



giving $\int_0^\infty xe^{-\alpha x} \ dx = \frac{1}{\alpha^2}$ (repeat for higher powers)

So this just simplifies something that you can already do by IBP or whatever. It also works fine for indefinite integrals - you don't need the limits.

or this for integrals which are hard:

2. Hard integrals:

$I(\alpha) = \int_0^\infty e^{-\alpha x} \frac{\sin x}{x} \ dx \Rightarrow I'(\alpha) = - \int_0^\infty e^{-\alpha x} \sin x \ dx = -\frac{1}{1+\alpha^2}$

so that

$I(\alpha) = C-\tan^{-1} \alpha$

Now put $\alpha=\infty \Rightarrow I(\alpha) = 0 \Rightarrow C=\pi/2$

so $I(\alpha) = \frac{\pi}{2}-\tan^{-1} \alpha$

Now put $\alpha=0 \Rightarrow \int_0^\infty \frac{\sin x}{x} \ dx = \frac{\pi}{2}$

There's a bit more to it than that (the limits can depend on the parameter too), but that's the rough idea. You take an ordinary integral, introduce a parameter $\alpha$ somewhere, then differentiate w.r.t to that parameter to play tricks as above. The difficulty is finding where to put $\alpha$. Note that I introduced that exponential function above, which helpfully goes to 0 and infinity at the ends of the interval, which remove it completely, giving the nice result at the end.

[The conditions, as I recall, are:

a) if you are integrating over a closed interval $[a,b]$ (which has a property called compactness), then you can do the above as long as both $f(x,\alpha), \frac{\partial (f(x,\alpha)}{\partial \alpha}$ are nice, smooth functions (i.e. continuous & differentiable)

b) if you are integrating over an infinite region like $[0, \infty)$ (which lacks the property of compactness), then you can do the above as long as property a) above holds, and that an integrable function $g(x)$ exists that dominates $f(x)$ i.e. that $f(x,\alpha) \le g(x)$ everywhere.

(Maybe someone clever can correct this if necessary: paging M. LoTF)]

Having read that, I'd be interested to see how you are intending to do this - I suspect that it may differ from my approach.

Wow! Brilliant explanation. Thank you for such an insightful reply.

My pleasure. I'll put up a couple of practise questions for you later.

Yes, please do! I'm not very good with LaTeX though, so bear with me when it comes to posting my attempts/solutions. Thanks once again.

Yes, please do! I'm not very good with LaTeX though, so bear with me when it comes to posting my attempts/solutions. Thanks once again.

Just write them on paper and post it. I have never ever used latex, too complicated, just about handle spoilers me.


Posted from TSR Mobile

OK, here are two of the "easy" variants. The first is very easy, the second is harder:

1. Evaluate $\int \cos\alpha nx \ dx$. Hence evaluate $\int x \sin nx \ dx$

2. Evaluate $\displaystyle \int \frac{\sin 2x}{1 + \sin^4 x} \ dx$. Hence evaluate $\displaystyle \int \frac{\sin 2x}{(1 + \sin^4 x)^2} \ dx$.

Hints for 2nd:

Just write them on paper and post it. I have never ever used latex, too complicated, just about handle spoilers me.


Posted from TSR Mobile

Wait till you get to university; can't avoid TeX forever

I have withdrawn from my Firm and Insurance. Hope ur happy Latex.

You go boy!!! You're really socking it to The Man there, what with their fancy college kid "mark up" languages, and la-di-da "maths formatting"!! Ha ha, that'll teach 'em!!

And you know what's even better? - it frees you up to spend time learning stuff that will be, like, *really* useful to you, such as the phrases:

"Would you like fries with that?"
"Have a nice da-a-aa-y!"
"You forgot your change, sir!"

not to forget the ever useful

"Can I have another shift, mate? I got bills to pay.."
"Spare a coupla coppers for an old ex-mathmo who's down on his luck, pal?"

Evaluate

LOOOOOOL!! No damn way!!!

LOOOOOOL!! No damn way!!!

It's such a cool identity though, I'm still marvelling at how on earth it holds.

haha yea it is nice. u derived it I assume? It seems a little odd cus from the denom. id guess its < than the actual integral if I hadn't seen it before.

Yeah, the derivation is surprisingly accessible and short. I swear, that word play tho. "odd cus from the denom"

Yeah, the derivation is surprisingly accessible and short. I swear, that word play tho. "odd cus from the denom"