For the discussion of math. Duh.

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andy11235
Posts: 38
Joined: Thu Oct 16, 2008 6:21 am UTC

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

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
Posts: 7
Joined: Sun May 17, 2009 9:47 pm UTC

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

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
Posts: 14
Joined: Wed Sep 24, 2008 4:48 am UTC

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

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.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

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

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

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.)
wee free kings

mr-mitch
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

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

$\int \sqrt{x+k \over x} dx$

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

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

t0rajir0u
Posts: 1178
Joined: Wed Apr 16, 2008 12:52 am UTC
Location: Cambridge, MA
Contact:

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

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
$dx = \frac{-2ky}{(y^2 - 1)^2} \, dy$
hence
$\int y \, dx = \int \frac{-2k y^2}{(y^2 - 1)^2} \, dy.$
Now a routine computation with partial fractions gives the answer as
$\frac{k}{2} \left( \frac{2y}{y^2 - 1} + \log \frac{y+1}{y-1} \right)$
and we substitute.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

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

$\int \sqrt{x+k \over x} dx$

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
$\frac{2k}{\sqrt{1 + \frac{k}{x}} - 1} - 2k\left(\sqrt{1 + \frac{k}{x}} - 1\right) - 4k \ln\left(\sqrt{1 + \frac{k}{x}} - 1\right) + C$

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]
wee free kings

Why Two Kay
Posts: 266
Joined: Sun Mar 23, 2008 6:25 pm UTC
Location: Plano, TX
Contact:

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

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.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

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

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:
$\int\sqrt{\frac{x + k}{x}} dx = \int\sqrt{1 + \frac{k}{x}}dx$
[imath]r = \frac{k}{x} \Longrightarrow x = \frac{k}{r} \Longrightarrow dx = -k\frac{dr}{r^2}[/imath]
$\int\sqrt{1 + r}\frac{-k}{r^2}dr$
[imath]u = 1 + r \Longrightarrow r = u - 1 \Longrightarrow dr = du[/imath] and [imath]r^2 = \left(u - 1\right)^2[/imath]
$-k\int\frac{\sqrt{u}}{\left(u-1\right)^2}du$
[imath]v = \sqrt{u} \Longrightarrow u = v^2 \Longrightarrow du = 2v dv[/imath]
$-k\int\frac{v}{\left(v^2 - 1\right)^2}2vdv = -2k\int\left(\frac{v}{(v + 1)(v - 1)}\right)^2 dv$
[imath]\frac{v}{(v + 1)(v - 1)} = \frac{1}{2}\left(\frac{1}{v + 1} + \frac{1}{v - 1}\right)[/imath]
$\frac{-k}{2}\int\left(\frac{1}{v + 1} + \frac{1}{v - 1}\right)^2 dv$
$\frac{-k}{2}\int\left(\frac{1}{(v + 1)^2} + \frac{2}{(v + 1)(v - 1)} + \frac{1}{(v - 1)^2}\right) dv$
[imath]\frac{1}{(v + 1)(v - 1)} = \frac{1}{2}\left(\frac{1}{v - 1} - \frac{1}{v +1}\right)[/imath]
$\frac{-k}{2}\int\left(\frac{1}{(v + 1)^2} - \frac{1}{v + 1} + \frac{1}{v - 1} + \frac{1}{(v - 1)^2}\right) dv$
$\frac{-k}{2}\left(\frac{-1}{v + 1} - \ln(v + 1) + \ln(v - 1) + \frac{-1}{v - 1}\right) + C$
$\frac{-k}{2}\left(\frac{-2v}{v^2 - 1} - \ln\left(\frac{v + 1}{v - 1}\right)\right) + C$
[imath]v = \sqrt{1 + \frac{k}{x}}[/imath]
$x\sqrt{1 + \frac{k}{x}} + \frac{k}{2}\ln\left(1 + \frac{2x}{k}\left[1 + \sqrt{1 - \frac{k}{x}}\right]\right) + C$
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.
wee free kings

mr-mitch
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

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

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

Anywho, I award you your prize

Spoiler:
$i$

t0rajir0u
Posts: 1178
Joined: Wed Apr 16, 2008 12:52 am UTC
Location: Cambridge, MA
Contact:

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

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:
$\int_{0}^{1} \frac{\ln (x + 1)}{x^2 + 1} \, dx.$
(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.)

Afif_D
Posts: 184
Joined: Wed Oct 06, 2010 2:56 pm UTC

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

$\int \sqrt{x+k \over x} dx$

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)

Plasma_Wolf
Posts: 148
Joined: Mon Aug 22, 2011 8:11 pm UTC

### 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] )

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:
$\int \frac{1}{x \log x} \,dx$

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

We get:
$\int \frac{1}{u}du$

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
Posts: 221
Joined: Sun Dec 16, 2007 11:38 am UTC

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

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:
$\int \frac{1}{x \log x} \,dx$

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

We get:
$\int \frac{1}{u}du$

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.