Alrighty then. I think I'm muddying up a more straightforward method with unnecessary working but oh well. Also cba to include constants anywhere and I dunno why the formatting with dx and **** is going so mental

Spoiler

Reply 1387

SamDavies1998

4

What is DUTIS?

Reply 1388

username1162972

18

Original post by SamDavies1998

What is DUTIS?

Differenitiation under the integral sign. Consider a function of something(which isn't what your integrating wrt) then differentiate **** and get some easy integral then bang. Integerate vack to get normal ting and get the constant

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Reply 1389

Lord of the Flies

19

Original post by SamDavies1998

What is DUTIS?

Spoiler

^ motto of this thread

(edited 7 years ago)

Reply 1390

SamDavies1998

4

Original post by Lord of the Flies

Spoiler

^ motto of this thread

I'll try my best to remember that then.

Could you give me a very basic example of how this works, please? Or, alternatively, could you direct me to a source which explains this technique well. Thanks

Reply 1391

EnglishMuon

16

Original post by SamDavies1998

I'll try my best to remember that then.

Could you give me a very basic example of how this works, please? Or, alternatively, could you direct me to a source which explains this technique well. Thanks

Google is often handy with answering questions :smile:

:

https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

Reply 1392

atsruser

11

Original post by SamDavies1998

I'll try my best to remember that then.

Could you give me a very basic example of how this works, please? Or, alternatively, could you direct me to a source which explains this technique well. Thanks

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

Unparseable latex formula:

\frac{\partial}{\partial \alpha} \int_a^b f(x,\alpha)\ dx = \int_a^b \frac{\partial}{\partial \alpha} f(x,\alpha)\ dx }

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

1. Easy integrals:

Unparseable latex formula:

I(\alpha) = \int_0^\infty e^{-\alpha x} \ dx = \frac{1}\alpha}

so that

Unparseable latex formula:

I'(\alpha) = \int_0^\infty \frac{\partial}{\partial \alpha} e^{-\alpha x} \ dx = \frac{\partial}{\partial \alpha} \frac{1}\alpha}

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)]

(edited 7 years ago)

Reply 1393

Lord of the Flies

19

[The statement in that Wikipedia article is quite weak - below is a looser set of conditions]

Original post by SamDavies1998

...

Original post by atsruser

The conditions, as I recall, are [...]

It is the partial derivative that needs to be dominated, not

f

. Also you can merge a) and b) by dropping continuity in the first, since continuity on

[a,b]\times [\alpha,\beta]

implies boundedness, by compactness as you mentioned. So:

Suppose we are integrating on

X

, and that our parameter has domain

Y

. If there exists integrable

\Omega:X\to\mathbb{R}

such that

|\partial_{\alpha} f|\leqslant \Omega

on

X\times Y

(in fact we only need that to be true almost everywhere on

X

) then we can freely differentiate under the integral.

You can relax this domination condition a bit, though it becomes a bit awkward, and perhaps less everyday "useful". If for each

\alpha\in Y

there exists open

U_{\alpha}\ni\alpha

in

Y

and integrable

\Omega_{\alpha}:X\to\mathbb{R}

such that

|\partial_{\alpha} f|\leqslant \Omega_{\alpha}

on

X\times U_{\alpha}

then we can pass the derivative under the integral. In other words you don't need

f

to be dominated by a single function on the entirety of

X\times Y

.

[for anyone relatively new to this: I have assumed that

X,Y

are not pathological, so for the sake of this thread, interpret them as intervals (possibly including infinity)]

It should be noted that these are not necessary conditions: I skimmed through those a while ago, when I asked myself this question, and I don't remember what they were about - they are a fair bit more involved than any of this though, and most likely not relevant to the nature of this thread.

(edited 7 years ago)

Reply 1394

atsruser

11

Show that, for

a,b >0

:

\displaystyle \int_0^\infty \ln ( \frac{1-e^{bx}}{1-e^{ax}}) + (a-b)x \ dx = \frac{\pi^2}{6}(\frac{1}{a}-\frac{1}{b})

This surprised me when I derived it, but it appears to be correct. It looks like it may be divergent, but the log term tends asymptotically to cancel the linear term in just the right amount.

(edited 7 years ago)

Reply 1395

atsruser

11

Original post by atsruser

Show that, for

a > c

:

\displaystyle \int_0^{2\pi} \frac{\sin^{n-1} x \cos x(a \cos^{n-2} x+c)}{(a \cos^n x + c)^2 + b^2 \sin^{2n} x} \ dx = \frac{2\pi}{nb}

Hint:

Spoiler

Reply 1396

Zacken

22

Original post by atsruser

Show that [...] This surprised me when I derived it, but it appears to be correct.

Do you need any complex magicz for this one? The

\frac{\pi^2}{6}

is shouting at me to think of sums, but my mind is too groggy to do maths at this time.

Reply 1397

atsruser

11

Original post by Zacken

Do you need any complex magicz for this one? The

\frac{\pi^2}{6}

is shouting at me to think of sums, but my mind is too groggy to do maths at this time.

Well, I don't think that it can be done by elementary methods, since the log part of the integrand turns into a dilogarithm antiderivative. Or did you mean contour integration type stuff? If so, no, I didn't use that. This is within your abilities (if not your knowledge).

You are thinking along the right lines with sums, but there may be a twist. Look back at my recent previous integrals to see where sums arise.

Nostradamus type hint:

Spoiler

Reply 1398

Lord of the Flies

19

Original post by atsruser

Show that [...]

You might want to add some conditions on a & b :biggrin:

Original post by Zacken

Do you need any complex magicz for this one?

It is a one-liner, either quoting the famous sum or quoting the zeta-gamma relation you mentioned earlier.

Reply 1399

atsruser

11

Original post by Lord of the Flies

You might want to add some conditions on a & b :biggrin:

Yes, I'll add them. Thanks.

I'll meditate deeply on your corrected conditions for DUTIS, too. I really ought to try to look at a proof, but I'm too rusty to understand them these days.

It is a one-liner, either quoting the famous sum or quoting the zeta-gamma relation you mentioned earlier.

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.

The hard integral thread.

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