Post your interesting/challenging/fun integrals

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andy11235
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Re: Post your interesting/challenging/fun integrals

Postby andy11235 » Sat Jul 18, 2009 5:00 am UTC

EclipseEnigma wrote:
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)


I think you made a sign error in that.
Spoiler:
I got [imath]\int \frac{1}{x \log x} \,dx = 1 + \int \frac{1}{x \log x} \,dx[/imath] which does not provide a solution.

Tmabbbb
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Re: Post your interesting/challenging/fun integrals

Postby Tmabbbb » Fri Jul 24, 2009 4:15 am UTC

I am only in ninth grade, but I believe I may be of assistance in this problem. I am not aware of how to make the mathematical symbols with my keyboard, so I will just describe the procedure. I replace (log x) for u, and 1/x becomes du. The resulting integral is then 1/u du, which becomes log u. Then, the u is replaced by (log x), and the answer is apparent, log(log x) + C.

Excalibur0998
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Re: Post your interesting/challenging/fun integrals

Postby Excalibur0998 » Sun Jul 26, 2009 1:47 am UTC

t0rajir0u wrote: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]?


I put them into wolfram alpha, and they look very similar...but what does [imath]y^5 = 1 - x^7[/imath] have to do with [imath](1 - x)^7[/imath]? Anyway, I finally sort of gave up on this integral so I went to my dad and he showed me how to do it using
Spoiler:
the Beta function and the Gamma function,
but if there's another approach (graphical approach) then I'd be interested in hearing it.

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Qaanol
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Re: Post your interesting/challenging/fun integrals

Postby Qaanol » Sun Jul 26, 2009 2:20 am UTC

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


Excalibur0998 wrote:
t0rajir0u wrote: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]?


I put them into wolfram alpha, and they look very similar...but what does [imath]y^5 = 1 - x^7[/imath] have to do with [imath](1 - x)^7[/imath]? Anyway, I finally sort of gave up on this integral so I went to my dad and he showed me how to do it using
Spoiler:
the Beta function and the Gamma function,
but if there's another approach (graphical approach) then I'd be interested in hearing it.

I'm not sure where you see [imath](1 - x)^7[/imath].

In any event, the two functions t04ajir0u mentions are inverses of one another. And they both go through the points (0, 1) and (1, 0). Think about what change-of-variables you could employ (or what symmetry argument.)
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mr-mitch
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Re: Post your interesting/challenging/fun integrals

Postby mr-mitch » Mon Jul 27, 2009 7:57 am UTC

Test your brilliance:

[math]\int \sqrt{x+k \over x} dx[/math]

I will award one complex prize to the first person who solves it.

PS It's hell, although that's possibly an understatement.

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t0rajir0u
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Re: Post your interesting/challenging/fun integrals

Postby t0rajir0u » Mon Jul 27, 2009 5:40 pm UTC

Er... really? It looks pretty straightforward to me.

Spoiler:
Make the obvious substitution [imath]y^2 = 1 + \frac{k}{x}[/imath]. This gives [imath]2y \, dy = - \frac{k}{x^2} \, dx[/imath]. Since [imath]x = \frac{k}{y^2 - 1}[/imath], this gives
[math]dx = \frac{-2ky}{(y^2 - 1)^2} \, dy[/math]
hence
[math]\int y \, dx = \int \frac{-2k y^2}{(y^2 - 1)^2} \, dy.[/math]
Now a routine computation with partial fractions gives the answer as
[math]\frac{k}{2} \left( \frac{2y}{y^2 - 1} + \log \frac{y+1}{y-1} \right)[/math]
and we substitute.

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Qaanol
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Re: Post your interesting/challenging/fun integrals

Postby Qaanol » Mon Jul 27, 2009 6:02 pm UTC

mr-mitch wrote:Test your brilliance:

[math]\int \sqrt{x+k \over x} dx[/math]

I will award one complex prize to the first person who solves it.

PS It's hell, although that's possibly an understatement.

Spoiler:
I may have made a mistake, but I seem to have solved it using repeated u-substitution. My answer is
[math]\frac{2k}{\sqrt{1 + \frac{k}{x}} - 1} - 2k\left(\sqrt{1 + \frac{k}{x}} - 1\right) - 4k \ln\left(\sqrt{1 + \frac{k}{x}} - 1\right) + C[/math]

Note if k is a constant the central term's contribution of 2k can be clumped with C.

My substitutions were:
[imath]r = k/x \Longrightarrow dx = -k r^{-2} dr[/imath]
[imath]u = 1 + r \Longrightarrow dr = du[/imath]
[imath]v = u^{1/2} \Longrightarrow du = 2v dv[/imath]
[imath]w = v - 1 \Longrightarrow dv = dw[/imath]

So at the end [imath]w = \sqrt{1 + \frac{k}{x}} - 1[/imath]
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Why Two Kay
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Re: Post your interesting/challenging/fun integrals

Postby Why Two Kay » Mon Jul 27, 2009 6:55 pm UTC

Plug that one into Wolfram Alpha and click "Show steps"

http://www07.wolframalpha.com/input/?i= ... 9%2Fx%29dx

Now there's a complicated and roundabout way of doing it, despite being possibly the easiest for a computer to apply.
tl;dr - I said nothing important.

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Qaanol
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Re: Post your interesting/challenging/fun integrals

Postby Qaanol » Mon Jul 27, 2009 8:18 pm UTC

That is impressive. Now I realize I did make a mistake. t0's right, partial fractions are useful. Here's my corrected answer with steps:

Spoiler:
[math]\int\sqrt{\frac{x + k}{x}} dx = \int\sqrt{1 + \frac{k}{x}}dx[/math]
[imath]r = \frac{k}{x} \Longrightarrow x = \frac{k}{r} \Longrightarrow dx = -k\frac{dr}{r^2}[/imath]
[math]\int\sqrt{1 + r}\frac{-k}{r^2}dr[/math]
[imath]u = 1 + r \Longrightarrow r = u - 1 \Longrightarrow dr = du[/imath] and [imath]r^2 = \left(u - 1\right)^2[/imath]
[math]-k\int\frac{\sqrt{u}}{\left(u-1\right)^2}du[/math]
[imath]v = \sqrt{u} \Longrightarrow u = v^2 \Longrightarrow du = 2v dv[/imath]
[math]-k\int\frac{v}{\left(v^2 - 1\right)^2}2vdv = -2k\int\left(\frac{v}{(v + 1)(v - 1)}\right)^2 dv[/math]
[imath]\frac{v}{(v + 1)(v - 1)} = \frac{1}{2}\left(\frac{1}{v + 1} + \frac{1}{v - 1}\right)[/imath]
[math]\frac{-k}{2}\int\left(\frac{1}{v + 1} + \frac{1}{v - 1}\right)^2 dv[/math]
[math]\frac{-k}{2}\int\left(\frac{1}{(v + 1)^2} + \frac{2}{(v + 1)(v - 1)} + \frac{1}{(v - 1)^2}\right) dv[/math]
[imath]\frac{1}{(v + 1)(v - 1)} = \frac{1}{2}\left(\frac{1}{v - 1} - \frac{1}{v +1}\right)[/imath]
[math]\frac{-k}{2}\int\left(\frac{1}{(v + 1)^2} - \frac{1}{v + 1} + \frac{1}{v - 1} + \frac{1}{(v - 1)^2}\right) dv[/math]
[math]\frac{-k}{2}\left(\frac{-1}{v + 1} - \ln(v + 1) + \ln(v - 1) + \frac{-1}{v - 1}\right) + C[/math]
[math]\frac{-k}{2}\left(\frac{-2v}{v^2 - 1} - \ln\left(\frac{v + 1}{v - 1}\right)\right) + C[/math]
[imath]v = \sqrt{1 + \frac{k}{x}}[/imath]
[math]x\sqrt{1 + \frac{k}{x}} + \frac{k}{2}\ln\left(1 + \frac{2x}{k}\left[1 + \sqrt{1 - \frac{k}{x}}\right]\right) + C[/math]
For the last substitution, I simplified the first term by obvious means, and the argument to the log by multiplying and dividing by (v + 1).

Same as t0's.
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mr-mitch
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Re: Post your interesting/challenging/fun integrals

Postby mr-mitch » Tue Jul 28, 2009 1:10 am UTC

It takes quite a while to do, and there isn't some simple and brilliant subsitution (that I'm aware of) which solves it in a couple of lines... so that's hell :P

Anywho, I award you your prize

Spoiler:
[math]i[/math]

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t0rajir0u
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Re: Post your interesting/challenging/fun integrals

Postby t0rajir0u » Tue Jul 28, 2009 2:37 am UTC

mr-mitch wrote:It takes quite a while to do

Sure it's tedious, but it's far from "hell." It doesn't require any non-obvious steps. Everything was mechanical after the substitution and the substitution was obvious. Just because an integral is annoying to do by hand doesn't make it hard. On the other hand, here's a genuinely hard problem:
[math]\int_{0}^{1} \frac{\ln (x + 1)}{x^2 + 1} \, dx.[/math]
(Hard in the sense that some steps aren't obvious, at least if you're willing to do this without contour integration. Also hard in the sense that it was on a Putnam.)

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Afif_D
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Re: Post your interesting/challenging/fun integrals

Postby Afif_D » Thu Oct 13, 2011 11:41 am UTC

mr-mitch wrote:Test your brilliance:

[math]\int \sqrt{x+k \over x} dx[/math]

I will award one complex prize to the first person who solves it.

PS It's hell, although that's possibly an understatement.


Substituting x=k*((tan(z))^2)

makes the integral turn relatively simple...(Thats the clever and obvious substitution)
Image

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Plasma_Wolf
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Re: Post your interesting/challenging/fun integrals

Postby Plasma_Wolf » Sat Oct 15, 2011 5:22 pm UTC

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] )

I assume that the u has to be a v or the dv has to be an u.

The way I get to the answer:
Spoiler:
[math]\int \frac{1}{x \log x} \,dx[/math]

Substitue:
[imath]u=\log{x}[/imath] and [imath]du=\frac{1}{x}dx[/imath]

We get:
[math]\int \frac{1}{u}du[/math]

Integrate here and you get:
[imath]\log{u} +C[/imath]

Don't forget that you used a substitution:
[imath]\log{(\log{x})} +C[/imath]

Yesila
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Re: Post your interesting/challenging/fun integrals

Postby Yesila » Sun Oct 16, 2011 7:28 am UTC

Plasma_Wolf wrote:
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] )

I assume that the u has to be a v or the dv has to be an u.

The way I get to the answer:
Spoiler:
[math]\int \frac{1}{x \log x} \,dx[/math]

Substitue:
[imath]u=\log{x}[/imath] and [imath]du=\frac{1}{x}dx[/imath]

We get:
[math]\int \frac{1}{u}du[/math]

Integrate here and you get:
[imath]\log{u} +C[/imath]

Don't forget that you used a substitution:
[imath]\log{(\log{x})} +C[/imath]



I think the point was that even a standard substitution integral such as this can be "fun" by asking for it to be done by a different method.


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