For the discussion of math. Duh.

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oblivimous
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For many students, integration is an exciting topic because, especially at first, every problem is a puzzle. After years of doing mathematics where textbook-drill-problems can be solved by rote, suddenly problems require varying degrees of cleverness.

Let's collect the best here. Problems should be solvable with the background available to a typical first-year calculus student, but should require exceptional cleverness, nontraditional approaches, etc.

1) [imath]\int e^x \cos x \,dx[/imath]

Spoiler:
integration by parts twice

2) [imath]\int_0^1 \sqrt{4-x^2} \, dx[/imath]

Spoiler:
For a student who has not yet learned the formula for integration of a polar equation (f->r=2), this integral is best solved as an area problem

Ended
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### Re: Post your interesting/challenging/fun integrals

3) [imath]\int \ln(x)\ dx[/imath]

Spoiler:
Write ln(x) = 1*ln(x) and integrate by parts.

Edit: log->ln
Last edited by Ended on Sun Jul 12, 2009 5:13 pm UTC, edited 1 time in total.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
-dubsola

Torn Apart By Dingos
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### Re: Post your interesting/challenging/fun integrals

This is a classic:

4) [imath]\int_{-\infty}^\infty e^{-x^2} dx[/imath]
Spoiler:
This is the Gaussian integral. The really clever calculation can be found on Wikipedia, so I won't repeat it here.

Regarding 1):
Spoiler:
It can also be done by equating it with the real part of [imath]\int e^x e^{ix} dx[/imath]. One needs some complex analysis to justify this, but it's a slicker solution.

Birk
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### Re: Post your interesting/challenging/fun integrals

Ended wrote:3) [imath]\int \log(x)\ dx[/imath]

Spoiler:
Write log(x) = 1*log(x) and integrate by parts.

Spoiler:
[imath]xlog(x) - \frac{x}{ln10} ?[/imath]

Could stand to simplify it some more but that's what I got. It never occurred to me there isn't a rule to memorize for [imath]log(x)[/imath]

Edit:

Spoiler:
I shouldn't be so lazy.

$\frac{x(logx-1)}{ln10}$

Ended
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### Re: Post your interesting/challenging/fun integrals

Generally I try to make myself do things I instinctively avoid, in case they are awesome.
-dubsola

Birk
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Joined: Tue May 19, 2009 5:08 pm UTC

### Re: Post your interesting/challenging/fun integrals

Oh whoops just got sidetracked typing it out. Is using log(x) to mean ln(x) and then specifying log10(x) the norm in higher maths? Because I've noticed Wolfram assumes log(x) to be loge(x) where as I was always taught that log(x) is base 10 by it says otherwise.

Or was I taught wrong up through calc II?

NathanielJ
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### Re: Post your interesting/challenging/fun integrals

Birk wrote:Or was I taught wrong up through calc II?

In my experience, it depends on your background. I always think base-10 log when I see an unqualified log, like you. However, I have a pure math background, and that's the way they tend to do things in pure math courses (until you get to complex analysis, when log means the complex natural logarithm, which just messes everything up ). Computer scientists, in my experience, tend to use log to mean log2, and physicists tend to use log to mean loge (in my experience). There doesn't seem to be any great standard, because for 99% of purposes it's just a technicality and doesn't really matter (they all just differ by a constant).
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Ended
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### Re: Post your interesting/challenging/fun integrals

In my experience it's the other way around (pure mathematicians using log for loge and applied mathematicians using ln). Which I suppose just goes to show how arbitrary it is

Here's another neat logarithm one:

5) [imath]\int \frac{1}{x \ln(x) \ln^2(x) ... \ln^n(x)}\ dx[/imath]

where [imath]\ln^n(x) := \ln\ln...\ln(x)[/imath] (n times).

Spoiler:
Try the n=1 case (you should be able to write down the answer without any working), then the n=2 case (again no real working needed, assuming you've done the n=1 case). This should tell you what the answer is, and give you a very simple and elegant inductive argument for the general case.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
-dubsola

csrjjsmp
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### Re: Post your interesting/challenging/fun integrals

One of my favorites:
$\int_0^1(1-x^7)^{1/5} - (1-x^5)^{1/7}dx$

Guff
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Joined: Thu Jan 03, 2008 11:56 pm UTC

### Re: Post your interesting/challenging/fun integrals

Irresistible Integrals has some great ones.

One pretty integral is
$\int_0^1 \ln{\Gamma(x)} dx = \ln{\sqrt{2\pi}}$

But I think maybe that would be a little more involved than what you're looking for.
If they've dealt with infinite sums, then the Sophomore's dream might be more approachable:
$\int_0^1 x^{-x} dx = \sum_{k=1}^\infty k^{-k}$

Fluid_Dynamic
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### Re: Post your interesting/challenging/fun integrals

Here's one that I stumped my high school class with for a few days:

[imath]\int \frac{1}{x \log x} \,dx[/imath]

Try solving it through integration by parts ( [imath]u = \frac{1}{\log x}, dv = \frac{1}{x} \,dx[/imath] )

Blatm
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### Re: Post your interesting/challenging/fun integrals

Ended wrote:In my experience it's the other way around (pure mathematicians using log for loge and applied mathematicians using ln). Which I suppose just goes to show how arbitrary it is

Here's another neat logarithm one:

5) [imath]\int \frac{1}{x \ln(x) \ln^2(x) ... \ln^n(x)}\ dx[/imath]

where [imath]\ln^n(x) := \ln\ln...\ln(x)[/imath] (n times).

Spoiler:
Try the n=1 case (you should be able to write down the answer without any working), then the n=2 case (again no real working needed, assuming you've done the n=1 case). This should tell you what the answer is, and give you a very simple and elegant inductive argument for the general case.

What I like about this one is that it's one of the few cases where the antiderivative won't exist everywhere the integrand does, so you have to be a bit careful.

t0rajir0u
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### Re: Post your interesting/challenging/fun integrals

oblivimous wrote:2) [imath]\int_0^1 \sqrt{4-x^2} \, dx[/imath]

Spoiler:
For a student who has not yet learned the formula for integration of a polar equation (f->r=2), this integral is best solved as an area problem

There is a perfectly elementary way to compute the indefinite integral here; just slice up the corresponding area into a sector and a triangle. One obtains in this way the power series for the arcsine with almost no work.

There's also a blog called the Daily Integral that people in this thread might find interesting.

ShaiDeshe
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### Re: Post your interesting/challenging/fun integrals

oblivimous wrote:
1) [imath]\int e^x \cos x \,dx[/imath]

Spoiler:
[imath]\int e^x \cos x \,dx = Re \int e^{(1+i)x} \,dx = Re \frac {e^{(1+i)x}} {1+i} = Re \frac {e^x(cos x + i sin x)} {1+i} = ...[/imath]

achan1058
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### Re: Post your interesting/challenging/fun integrals

Birk wrote:Oh whoops just got sidetracked typing it out. Is using log(x) to mean ln(x) and then specifying log10(x) the norm in higher maths? Because I've noticed Wolfram assumes log(x) to be loge(x) where as I was always taught that log(x) is base 10 by it says otherwise.
It can be base 10, e, or 2 depending on your backgrounds. Personally, I use log for base 10, ln for base e, and lg for base 2.

Birk
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### Re: Post your interesting/challenging/fun integrals

achan1058 wrote:
Birk wrote:Oh whoops just got sidetracked typing it out. Is using log(x) to mean ln(x) and then specifying log10(x) the norm in higher maths? Because I've noticed Wolfram assumes log(x) to be loge(x) where as I was always taught that log(x) is base 10 by it says otherwise.
It can be base 10, e, or 2 depending on your backgrounds. Personally, I use log for base 10, ln for base e, and lg for base 2.

I'm the same way for log, ln, lg. I didn't realize it was so arbitrary

auteur52
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### Re: Post your interesting/challenging/fun integrals

If you're going to study pure math, get into the habit of using log for base-e. Almost every textbook will use this without noting what it means, as well as pretty much every single lecturer.

achan1058
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### Re: Post your interesting/challenging/fun integrals

auteur52 wrote:If you're going to study pure math, get into the habit of using log for base-e. Almost every textbook will use this without noting what it means, as well as pretty much every single lecturer.
Yes, but you see the problem is I study both math and CS, and in CS they always use log for base 2. I suspect it to get more fun in computation math, where when they show how to calculate log, it would be base e, while when they do the runtime analysis 5 minutes later, it will be base 2, even though it usually doesn't matter under the big-O unless it is at the exponent.

stephentyrone
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### Re: Post your interesting/challenging/fun integrals

achan1058 wrote:
auteur52 wrote:If you're going to study pure math, get into the habit of using log for base-e. Almost every textbook will use this without noting what it means, as well as pretty much every single lecturer.
Yes, but you see the problem is I study both math and CS, and in CS they always use log for base 2. I suspect it to get more fun in computation math, where when they show how to calculate log, it would be base e, while when they do the runtime analysis 5 minutes later, it will be base 2, even though it usually doesn't matter under the big-O unless it is at the exponent.

The names in the C standard math library say it all: log, log10, and log2. Computer scientists who use log and mean log-base-2 are either (a) taking advantage of the fact that constants don't matter or (b) lazy.
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mike-l
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### Re: Post your interesting/challenging/fun integrals

I use log and ln interchangeably both to mean base e.

My experience has been the same with basically every other mathematician I've worked with (though most of the time constants don't matter anyway, so it's kind of unimportant). If you want to talk about log base N, I'd suggest using ln x/ln N, since this is how it should be thought of in most cases.

NathanielJ
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### Re: Post your interesting/challenging/fun integrals

stephentyrone wrote:The names in the C standard math library say it all: log, log10, and log2. Computer scientists who use log and mean log-base-2 are either (a) taking advantage of the fact that constants don't matter or (b) lazy.

or (c) information theorists.
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Torn Apart By Dingos
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### Re: Post your interesting/challenging/fun integrals

csrjjsmp wrote:One of my favorites:
$\int_0^1(1-x^7)^{1/5} - (1-x^5)^{1/7}dx$

I assume this is supposed to be 0. Why?

t0rajir0u
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### Re: Post your interesting/challenging/fun integrals

What does the graph of [imath]y^5 = 1 - x^7[/imath] look like? How does it relate to the graph of [imath]y^7 = 1 - x^5[/imath]?

TheWaterBear
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### Re: Post your interesting/challenging/fun integrals

Personal favorite:

$\int_0^{ \pi\ /2} \ sinx/(sinx + cosx) \, dx = \pi\ /4$

Spoiler:
Note that over the given interval the function cosx/(sinx + cosx) attains the exact same values as sinx/(sinx + cosx), just in the opposite order.
Last edited by TheWaterBear on Tue Jul 14, 2009 3:27 am UTC, edited 1 time in total.

jestingrabbit
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### Re: Post your interesting/challenging/fun integrals

Does anyone else have a problem with the first four spoilers not opening?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

loophole21
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### Re: Post your interesting/challenging/fun integrals

Heres a tough antiderivative if you've never seen it before. It requires a clever substitution then partial fractions:

$\int \frac{sinx}{1+sinx}\ dx$

Spoiler:
Let z = tan(x/2) and derive expressions for sinx and dx

t0rajir0u
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### Re: Post your interesting/challenging/fun integrals

Do people really consider that substitution "clever"? If you've seen the geometric proof of the Pythagorean triple identities, it's the first thing that should come to mind. The substitution also shows that every rational function of trig functions has an elementary antiderivative.

jestingrabbit
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### Re: Post your interesting/challenging/fun integrals

I also immeadiately thought of that substitution. Its the usual "I have an integral involving trig functions and no idea how to make it work" substitution.

Noone else have a problem with the first four spoilers not opening?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

Why Two Kay
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### Re: Post your interesting/challenging/fun integrals

jestingrabbit wrote:Noone else have a problem with the first four spoilers not opening?

They don't open for me in Firefox. I thought it was just me.
tl;dr - I said nothing important.

LLCoolDave
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### Re: Post your interesting/challenging/fun integrals

It's clever in as so far as that it is not obvious the first time you see it. If you've used it before I'm fairly certain your view is somewhat biased because it's a useful substitution to remember.

Here's one our class did recently (It came up in the context of balckbody radiation, which in itsself was already far from the subject matter. This class has gone so astray I don't even remember its goal anymore.)

$\int_{0}^{\infty} \frac{x^3}{e^x-1} dx = \frac{\pi^4}{15}$

Spoiler:
The Route we took was to artificially rewrite the denominator as a geometric Series over [imath]e^{-kx}[/imath], then use some theorems in order to pull the sum outside the (indefinite) Integral, evaluating the integral by Integration by parts and finally evaluating the infinite series using a Fourierseries. Took well over an hour just to present it, but it was interesting because it doesn't look all that bad to solve this beast. There may be more straight forward approaches, but I am not too certain how much more reasonable they are.

skeptical scientist
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### Re: Post your interesting/challenging/fun integrals

I'm having no issues opening spoilers in Safari. I'm also running subsilver instead of prosilver - I'm not sure if that matters.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

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t0rajir0u
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### Re: Post your interesting/challenging/fun integrals

LLCoolDave wrote:The Route we took was to artificially rewrite the denominator as a geometric Series over [imath]e^{-kx}[/imath], then use some theorems in order to pull the sum outside the (indefinite) Integral, evaluating the integral by Integration by parts and finally evaluating the infinite series using a Fourierseries.

Can you simplify the argument by showing that the integral is equal to [imath]\displaystyle 6 \sum_{k=1}^{\infty} \frac{1}{k^4}[/imath], by any chance? It seems like a natural thing to try given the value of the RHS.

I also don't see why the geometric series argument is "artificial." It's more or less the only way to prove the sophomore's dream, for example, and it's a good (if heavy-handed) technique to keep in mind.
Last edited by t0rajir0u on Tue Jul 14, 2009 6:43 am UTC, edited 1 time in total.

jestingrabbit
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### Re: Post your interesting/challenging/fun integrals

might just be firefox.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jaap
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### Re: Post your interesting/challenging/fun integrals

jestingrabbit wrote:might just be firefox.

I use Firefox 3.5 and the spoilers all open fine for me.

t0rajir0u
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### Re: Post your interesting/challenging/fun integrals

LLCoolDave wrote:$\int_{0}^{\infty} \frac{x^3}{e^x-1} dx = \frac{\pi^4}{15}$

On second thought, probably the most natural way to do this is contour integration, since this identity seems closely related to a connection between the Bernoulli numbers, whose generating function is [imath]\frac{x}{e^x - 1}[/imath], and the values of the zeta function at the even integers.

Edit: Define
$F(s) = \int_{0}^{\infty} \frac{te^{-st}}{1 - e^{-t}} \, dt$
i.e. the Laplace transform of [imath]\frac{t}{1 - e^{-t}}[/imath]. Since the Laplace transform of [imath]te^{-kt}[/imath] is equal to [imath]\frac{1}{(s+k)^2}[/imath] it follows that, after expanding the denominator of the integrand as a geometric series, the above is equal to
$F(s) = \sum_{k=0}^{\infty} \frac{1}{(s+k)^2}.$
Setting [imath]s = 1[/imath] we obtain
$\int_{0}^{\infty} \frac{t}{e^t - 1} \, dt = \zeta(2) = \frac{\pi^2}{6}.$
Taking two more derivatives we obtain
$F^{(2)}(s) = \sum_{k=0}^{\infty} \frac{6}{(s+k)^4}$
and setting [imath]s = 1[/imath] we obtain
$\int_{0}^{\infty} \frac{t^3}{e^t - 1} \, dt = 6 \zeta(4) = \frac{\pi^4}{15}.$
The generalization is clear. In fact it may even be the case that
$\int_{0}^{\infty} \frac{t^a}{e^t - 1} \, dt = \Gamma(a+1) \zeta(a+1).$

Qaanol
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### Re: Post your interesting/challenging/fun integrals

loophole21 wrote:Heres a tough antiderivative if you've never seen it before. It requires a clever substitution then partial fractions:

$\int \frac{sinx}{1+sinx}\ dx$

Spoiler:
Let z = tan(x/2) and derive expressions for sinx and dx

My first thought was
Spoiler:
Multiply and divide by [imath]1 - sinx[/imath].
$\int \frac{sinx}{1+sinx}\ dx = \int \frac{sinx}{1+sinx}\ \frac{1 - sinx}{1 - sinx}\ dx$
$= \int \frac{sinx - sin^2x}{1 - sinx^2}\ dx = \int \frac{sinx - sin^2x}{cos^2x}\ dx$
$= \int \frac{sinx}{cos^2x}\ dx +\int \frac{-sin^2x}{cos^2x}\ dx = \int secx\, tanx\, dx + \int \frac{1 - 1 - sin^2x}{cos^2x}\ dx$
$= \int secx\, tanx\, dx + \int \frac{-1}{cos^2x}\ dx + \int \frac{1 - sin^2x}{cos^2x}\ dx$
$= \int secx\, tanx\, dx - \int sec^2x\, dx + \int \frac{cos^2x}{cos^2x}\ dx$
$= \int secx\, tanx\, dx - \int sec^2x\, dx + \int dx$
$= secx - tanx + x + C = \frac{1 - sinx + x\, cosx}{cosx}\ + C$
It's true I introduced a discontinuity at [imath]x = \frac{3\pi}{2}\ + 2n\pi[/imath], but it's removable.
wee free kings

Guff
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### Re: Post your interesting/challenging/fun integrals

t0rajir0u wrote:Can you simplify the argument by showing that the integral is equal to [imath]\displaystyle 6 \sum_{k=1}^{\infty} \frac{1}{k^4}[/imath], by any chance?

Well, yeah, that is pretty straightforward, given that
Spoiler:
$\zeta(s) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}}{e^t - 1}dt$

which is straightforward to prove.

mr-mitch
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### Re: Post your interesting/challenging/fun integrals

Qaanol wrote:
loophole21 wrote:Heres a tough antiderivative if you've never seen it before. It requires a clever substitution then partial fractions:

$\int \frac{sinx}{1+sinx}\ dx$

Spoiler:
Let z = tan(x/2) and derive expressions for sinx and dx

My first thought was
Spoiler:
Multiply and divide by [imath]1 - sinx[/imath].
$\int \frac{sinx}{1+sinx}\ dx = \int \frac{sinx}{1+sinx}\ \frac{1 - sinx}{1 - sinx}\ dx$
$= \int \frac{sinx - sin^2x}{1 - sinx^2}\ dx = \int \frac{sinx - sin^2x}{cos^2x}\ dx$
$= \int \frac{sinx}{cos^2x}\ dx +\int \frac{-sin^2x}{cos^2x}\ dx = \int secx\, tanx\, dx + \int \frac{1 - 1 - sin^2x}{cos^2x}\ dx$
$= \int secx\, tanx\, dx + \int \frac{-1}{cos^2x}\ dx + \int \frac{1 - sin^2x}{cos^2x}\ dx$
$= \int secx\, tanx\, dx - \int sec^2x\, dx + \int \frac{cos^2x}{cos^2x}\ dx$
$= \int secx\, tanx\, dx - \int sec^2x\, dx + \int dx$
$= secx - tanx + x + C = \frac{1 - sinx + x\, cosx}{cosx}\ + C$
It's true I introduced a discontinuity at [imath]x = \frac{3\pi}{2}\ + 2n\pi[/imath], but it's removable.

I used
Spoiler:
long division,

1 - 1/(1+sin x)
'simplifying' the integral (by rationalisation) to the integral of

1 - (1-sinx)/cos²(x)
or
1 - sec²x + sec x tan x

which is very simply,

x - tan x + sec x + C

I like the integral of cosecant (i.e. 1/sin x)

EclipseEnigma
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### Re: Post your interesting/challenging/fun integrals

Fluid_Dynamic wrote:Here's one that I stumped my high school class with for a few days:

[imath]\int \frac{1}{x \log x} \,dx[/imath]

Try solving it through integration by parts ( [imath]u = \frac{1}{\log x}, dv = \frac{1}{x} \,dx[/imath] )

Spoiler:
Unless I am doing something wrong, I get
[imath]\int \frac{1}{x \log x} \,dx = 1 - \int \frac{1}{x \log x} \,dx[/imath]
So [imath]\int \frac{1}{x \log x} \,dx = 1/2[/imath]

Although the answer should clearly be log(log x)

Gnurdux
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### Re: Post your interesting/challenging/fun integrals

t0rajir0u wrote:
LLCoolDave wrote:$\int_{0}^{\infty} \frac{x^3}{e^x-1} dx = \frac{\pi^4}{15}$

The generalization is clear. In fact it may even be the case that
$\int_{0}^{\infty} \frac{t^a}{e^t - 1} \, dt = \Gamma(a+1) \zeta(a+1).$

In fact, not only is this true, it is used in deriving the functional equation for zeta. Also, he did mention reducing it to an infinite sum using a geometric series expansion, which would clearly be the sum of [imath]1/n^4[/imath].

Edit: The way to show this is to look at integrals of [imath]e^{-kx}x^s[/imath] and to note that [imath]\frac{1}{e^x-1}[/imath] is the sum of a geometric series. It turns out that it is kosher to swap the order of the improper integral and infinite sums, which you can show by explicitly looking at error terms.