Are You Ready to Challenge Your Integral Solving Skills?

[Gib Z]

Hey guys! Some people in the homework section wanted this thread revived, but with a new intent.

The original purpose was for people to post up integrals (usually indefinite) for me to solve. However the renewed purpose is for anyone to post up a particularly difficult problem for anyone to solve. The problems should be able to be worked out with no more than CalcII knowledge please. It does not necessarily have to be an integral, it could be an interesting piece of number theory or the computation of the value of a series, or even a particularly difficult exercise of algebraic manipulation!

It would be wonderful if you all would participate :)

It has concerned some that this thread will be exploited by some for their homework to be solved, and a way to avoid this would be extra work on part of the mentors. For this I propose that Only members with post counts of over 500, or members who post solutions to the previous problem, will be allowed to post new questions. Generally I have found people with higher post counts know quite a lot and are more accustomed to the forums rules, so will not exploit this. And granted members can solve the previous problem, they probably would not need help with their homework :)

I think I will go first :) (My Class is learning the compound interest formula, this is definitely not my homework):

Evaluate \sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^m\left(m\,3^n+n\,3^m\right)}

This is quite hard.

 

[mathwonk]

how about problems for which one needs to set up the integral, like say, find the "volume" of the unit ball in 4 space?

(by the way this thread title reminds me of the old song "how dry i am")

 

[Parthalan]

Any excuse to use hyperspherical coordinates (I think) will do!

 

[Gib Z]

Hyperspherical co ordinates and 4 space doesn't sound like CalcII level to me :( I guess if you wanted you could post such questions, but I definitely would not be able to participate :P

 

[yip]

In response to GibZ's symmetrical question:
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^ m\left(m\,3^n+n\,3^m\right)}=\frac{1}{2}[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^ m\left(m\,3^n+n\,3^m\right)}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{n^2\,m}{3^ n\left(n\,3^m+m\,3^n\right)}]
=\frac{1}{2}[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^{2}n.3^{n}(n3^{m}+m3^{n})+n^{2}m.3^{m}(m3^{n}+n3^{m})}{3^{m+n}(m3^{n}+n3^{m})^{2}}]
=\frac{1}{2}[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{mn}{3^{m+n}}]=\frac{1}{2}\sum_{m=1}^\infty\frac{m}{3^{m}}\sum_{n=1}^\infty\frac{n}{3^{n}}
\sum_{m=1}^\infty\frac{m^{2}}{3^{m}}=(\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+...)
=(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...)+(\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...)+(\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+...)+...
=\frac{\frac{1}{3}+\frac{1}{9}+...}{\frac{2}{3}}=\frac{3}{4}
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^ m\left(m\,3^n+n\,3^m\right)}=\frac{1}{2}\frac{9}{16}=\frac{9}{32}

 

[Gib Z]

yip, you are one smart cookie :) That is exactly correct, which means you get to post the next question if you want :)

I didn't think anyone would solve it that fast >.<

 

[yip]

Here is a rather tricky integral:

Integrate

\int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}

 

[Gib Z]

Mathematica gives the answer is terms of various non-elementary functions...But I've done it before and I can do it again, I'm sure its much simpler than the Integrator says it is.

 

[Gib Z]

\int{\frac{{(1+x^{2})}}{{(1-x^{2})\sqrt{1+x^{4}}}}dx = }\int{\frac{{x^{2}(\frac{1}{{x^{2}}}+1)}}{{x^{2}(\frac{1}{x}-x)\sqrt{\frac{1}{{x^{2}}}+x^{2}}}}dx}

Let {(\frac{1}{x}-x)}=t, then {(\frac{1}{{x^{2}}}+1)dx}=-dt.

\int{\frac{{(1+x^{2})}}{{(1-x^{2})\sqrt{1+x^{4}}}}dx = }-\int{\frac{{dt}}{{t\sqrt{t^{2}+2}}}} = -\log \left({\frac{t}{{\sqrt{t^{2}+2}+\sqrt 2 }}}\right) + C

Rewrite in terms of x and we are done :)

 

[yip]

That seems to be correct, though I used the substitution t=x/(1-x^2), which works out a bit nicer

 

[Gib Z]

Yay :) So then its my turn? Try:

\int_{0}^{\infty}\;\frac{2\;-\;2\cos{x}}{x\;e^{x}}\;dx\;=\;\ln{2}

 

[yip]

Let

u(t)=\int_{0}^{\infty}\;\frac{2\;-\;2\cos{x}}{x\;e^{tx}}\;dx\;
Differentiating under the integral with respect to t,
\frac{du}{dt}=-\int_{0}^{\infty}\frac{(2-2cosx)dx}{e^{tx}}=\frac{2t}{t^{2}+1}-\frac{2}{t}
(For brevity I have omitted the working for du/dt, all I did was split the fraction, the first part is rather elementary, the second part can be found by 2 applications of integration by parts.)
u=log\frac{t^{2}+1}{t^{2}}+C where C is some constant
As t approaches infinity, u approaches 0, so C=0. Substitution of t=1 yields log2 for the integral desired.

 

[Gib Z]

Omg >.< These are ment to take several days for people to solve! Obviously they are not hard enough..or yip is just a friggin genious, and I very much suspect. They solution is largely correct but could you justify the differentiation under the integral sign?

 

[yip]

http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
Its quite a handy trick for solving strange integrals, ie this technique can be used to prove the famous property that the integral from 0 to infinity of sinx/x equals pi/2, which is rather nifty.

 

[Gib Z]

Ok then..You can post up another question then lol, while I try proving the integral you just said.

 

[mathwonk]

the method of volumes by slicing says that if you know the area of every circular slice of a sphere, you can integrate them to get the volume of the sphere.

if you understand this idea, then you will understand that knowing the volume of all the spherical slices of a 4 ball let's you integrate to get the 4 diml volume . this has nothing to do with exotic coordinates. (not that there's anything wrong with those.)

 

[JohnDuck]

I already solved mathwonk's question when he posed it in a different thread some time ago, so I'll leave that to someone else. I was, however, intrigued by this differentiating under the integral technique (which I've never encountered before). I went ahead and evaluated \int_{0}^{\infty} \frac{\sin{x}}{x} dx as per yip's suggestion and decided it was neat enough to post. Hope other people think so as well.

After a bit of trial, I defined:

u:\mathbb{R}^{+} \rightarrow \mathbb{R}

u(t)=\int_{0}^{\infty} \frac{e^{-tx}\sin{x}}{x} dx

Chosen because the limit as t approaches zero is \int_{0}^{\infty} \frac{\sin{x}}{x} dx and e^{-tx} approaches zero as x approaches infinity.

\frac{du}{dt} = -\int_{0}^{\infty} e^{-tx}\sin{x}dx

Which after a couple of applications of integration by parts becomes:

-\int_{0}^{\infty} e^{-tx}\sin{x}dx \ = \ \frac{e^{-tx}\sin{x}}{t} \ + \ \frac{e^{-tx}\cos{x}}{t^{2}} \ + \ \frac{1}{t^{2}} \int_{0}^{\infty} e^{-tx}\sin{x}dx

Evaluate the terms outside of integrals (the first becomes zero and the second becomes \frac{-1}{t^{2}}) and rearrange to get:

\frac{du}{dt} = \frac{-1}{t^{2} + 1} \ \Rightarrow \ u = -\arctan{t} + k

Where k is a constant. The limit of u as t approaches infinity is zero (can be seen from the original definition), which implies k is \frac{\pi}{2}. Take the limit of u as t approaches zero to find that:

\int_{0}^{\infty} \frac{\sin{x}}{x} = \frac{\pi}{2}

 

[Gib Z]

Very Nice Proof there JohnDuck, I haven't countered the technique previously either, hence my initial confusion at yip's proof and a request for justification. Good Work :)

And Mathwonk, I'm going to have to do some reading because I have no idea what a 4 ball is ...>.< Is it a 4 dimensional sphere?.. If so, this page helps me abit: http://www.mathpages.com/home/kmath163.htm .

(1/2) pi^2 R^4 , is that right?

 

[cliowa]

the method of volumes by slicing

Isn't that also referred to as "Cavalieris Principle"?
In your case, it would amount to the following, am I right?

volume of the 3-sphere=
2\cdot \int_{0}^{1}vol_3(S^2(\sqrt{1-r^2})) dr=2\cdot \int_{0}^{1}(\sqrt{1-r^2})^3 vol_3(S^2(1)) dr=\frac{8\pi}{3}\int_0^{1}(\sqrt{1-r^2})^3=\frac{\pi^2}{2},
where I made a trigonometric substition and some integration by parts to obtain the last equality.

 

[Gib Z]

Well It says on Wolfram: http://mathworld.wolfram.com/CavalierisPrinciple.html
that If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Is it just me, or is that oddly similar to the Theorems of Pappus? I find them amazingly useful for integrals that otherwise would be much harder.

Anyway, the volume of a Regular sphere is easy to derive! Its this 4 ball thing mathwonk talks about that confuses me :(

EDIT: I have the link for the other theorem that I was thinking of: http://mathworld.wolfram.com/PappussCentroidTheorem.html.

I think that The Principle is a generalization of the Centroid Theorem, or one of them is a special case of the other or something like that >.<

 

[cliowa]

Anyway, the volume of a Regular sphere is easy to derive! Its this 4 ball thing mathwonk talks about that confuses me :(

The 3-sphere is the 4 ball mathwonk is talking about!

It seems though that there are several things called Cavalieri's Principle, and I was not referring to the one you stated, GibZ. Maybe "volume by parallel sections" is a more popular name...

 

[Gib Z]

What is the 4 thing then >.<...:(

 

[cliowa]

What is the 4 thing then >.<...:(

Huh? In his original post mathwonk wrote "find the "volume" of the unit ball in 4 space". The unit ball in 4 space is the unit ball in 4 dimensional (real) space and that is the same thing as the 3-sphere (the dimension of the surface of this thing is 3, that's why it's called the 3-sphere, just like the 2-sphere S^2 lives in 3-dimensional space).

 

[mathwonk]

well after 30-40 years of teaching this stuff it dawned on me that the same idea we use to teach the volume of a ball ion 3 space actually the volume inductively, or recursively, of any ball in any space, and that we should teach this if we expect people to elarn ideas and not just formulas. cliowa has the computation for a unit ball and then using homogeneity the formula in gibz's post must be that for a general ball.

 

[mathwonk]

here is a site where someone else has done it:

http://alumni.umbc.edu/~ajohns5/4-sphere/4-sphere.html

 

[Gib Z]

I see >.< That is quite interesting, though maybe a bit out of my reach for now. And If yip doesn't return to this thread within 24 hours of my post, he will forfeit his right to post a question and it will go to the next person who wants it (and has at least 500 posts, for compliance reasons with PF rules).

 

[George Jones]

\sqrt{\pi} = \int_{-\infty}^{\infty}e^{-x^2}dx

can be used with the method of spherical shells to write down a closed-form expression for the volume of an n-sphere or n-ball in terms of the Gamma function.

I saw this in an introductory string theory book that I was reading a few years ago.

 

[yip]

New question as requested:

If an operator \Delta applied to a function of a has the effect of changing the a to a+1, and then subtracting the old function from this new function, show that

\Delta\int_{c}^{d}f(x,a)dx=\int_{c}^{d}\Delta f(x,a)dx
where c and d are independent of a.
Hence or otherwise solve
\int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{n}dx

 

[cliowa]

Maybe I didn't get this right; it seems trivial to me. Define a function g by g(a):=\int_{c}^{d}f(x,a)dx, then \Delta g(a)=g(a+1)-g(a)=\int_{c}^{d}f(x,a+1)dx-\int_{c}^{d}f(x,a)dx=\int_{c}^{d}f(x,a+1)-f(x,a)dx=\int_{c}^{d}\Delta f(x,a)dx, because surely f(x,a) is also a function of a (and assuming that f is nice enough).

 

[yip]

Yes, that is correct, the first part is rather simple. Its the second part that is very nice though, its not too hard, but I think it is quite quirky how the operator can help in evaluating certain integrals.

 

[Gib Z]

\int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{n}dx =(-1)^{n}\int_{0}^{\infty}e^{-a x}(1-e^{-x})^{n}\hspace{1 mm}dx

Letting u=1-e^{-x} we get:

(-1)^{n}\int_{0}^{1}(1-u)^{a-1}u^{n}\hspace{1 mm}du

For anyone familiar with the Beta or Gamma functions, it immediately evaluates to:

(-1)^{n}\beta(n+1,a)=\frac{(-1)^{n}\Gamma(n+1) \Gamma(a)}{\Gamma(n+a+1)}.

However for those who aren't, Repeated Integration by parts gives:
(-1)^n \frac{n! (a-1)!}{(a+n)!}.

 

[siddharth]

Rather simple, but I liked this one

\int \frac{\sin \theta - \cos \theta}{(\sin \theta + \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta

 

[yip]

Yes that is correct Gib Z, but that defeats the whole interstingness of the question. It is more interesting to apply the operator to the integral in question:

Since we may move the operator inside the integral as shown earlier,
\Delta\int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{n}dx=\int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{n+1}dx
Now let n=0
\Delta\int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{0}dx=\Delta\frac{1}{a}=\frac{1}{a+1}-\frac{1}{a}=-\frac{1}{a(a+1)}=\int_{0}^{\infty}e^{-ax}(e^{-x}-1)dx
\Delta\int_{0}^{\infty}e^{-ax}(e^{-x}-1)=\Delta-\frac{1}{a(a+1)}=\frac{1}{(a+1)(a+2)}+\frac{1}{a(a+1)}
=\frac{(-1)^{2}2!}{a(a+1)(a+2)}=\int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{2}
...<br /> \int_{0}^{\infty}e^{-ax}(e^{-x}-1)^{n}dx=\frac{(-1)^{n}n!}{a(a+1)...(a+n)}
I think that this operator approach is rather nice, I haven't seen this sort of approach to an integral before, which is why I had put this question up.

 

[Gib Z]

Siddarth, no ones gettting it. Want to post a hint please?

 

[ansrivas]

I=\int \frac{\sin \theta - \cos \theta}{(\sin \theta + \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta
= \int \frac{\sin^2 \theta - \cos^2 \theta}{(1 + 2\sin \theta \cos \theta)\sqrt{\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta}} d \theta

Now let u=\sin \theta \cos \theta
\frac{du}{d\theta}=\cos^2\theta-\sin^2\theta

This gives

I=\int \frac{-1}{(1+2u)\sqrt{u+u^2}}\,du

Now let l=\sqrt{u+u^2}
\frac{dl}{du}=\frac{1+2u}{2\sqrt{u+u^2}}

This gives
I=\int \frac{-2}{4l^2+1}dl

 

[mathwonk]

how about a 4 dimensional cone? if a 3 dimensional cone is given by

z = x^2 + y^2, with 0 <z <h, (i mean less than or equal)

then perhaps a 4 diml cone is given by w = x^2 + y^2 + z^2, with
0 < w < H.

is it also 1/3 the height times the base? or is it 1/4? or something else?

 

[mathwonk]

or a 4 diml tetrahedron. this would be a symmetric figure in 4 space whose faces are all 3 dimensional tetrahedra.

how many faces would it have?

or a 4 dimensional doughnut, obtained by revolving a 3 dimensional doughnut in x,y,z, space, around the y,z plane? one must fix radii for the three circles, subject to some simple inequalities.

 

[mathwonk]

heres an easier one: try to prove the area of a sphere is 4pi r^2 by first showing the sphere has the same area as the lateral area of a cylinder circumscribed about it. this appears in an 8th grade geometry book i am teaching out of next semester.

of course you must give some limit definition of the area of the sphere. but no calculus is needed, just geometry and limits. (i.e. precalculus.)

 

[rocomath]

now that I've had a little more calculus exposure, I'm ready to tackle this thread!

i remember when i first joined this forum, this thread made me **** my pants :-x

edit: i have 2 index cards worth of hard integrals, wish me luck! lol.

 

[rocomath]

Here's a slightly challenging one, it isn't too difficult, but not really simple either.

\int sec^3 x \ dx

this one is actually very ez

try \int\sec^{5}xdx

it's just lengthy and includes \int\sec^{3}xdx

 

[Gib Z]

I an really not prepared to do the whole thing, I will leave it as an exercise to you, however this method is much easier to follow than yours (no offense intended):

\int \sec^5 x dx = \int (1+\tan^2 x)\sqrt{1+\tan^2 x} \cdot \sec^2 x dx.

Let u= tan x.

Then
\int (1+u^2)^{3/2} du

Now let u= sinh y. Then:
\int \cosh^4 y dy

Using the squared to double angle identity twice gives the required result.

 

[rocomath]

omg! WOWWW, damn Gib Z. simply amazing.

don't worry about offending me! there's no harm in being shown a diff. method.

 

[Gib Z]

Now calm down, I don't deserve so much praise, although your last post has indeed reminded me of a comment that many people much more experienced in this than me, and I believe I should pass on to you: Don't worry if you can't always find the easiest method or any method of integration. You will learn with experience, somewhat by instinct, which methods will work and which substitutions will work the best.

One of the people who has told me that is dextercioby, who isn't as regular on the forums as he used to be, but he is truly a magician in the way he solved many integrals, mind boggling substitutions that amazingly simplified the problem or an expression of an integrand in a certain special way (like what I just did, though usually more ingenious). When I was a newbie on the forums I would always wonder why he would always prefer a hyperbolic trig substitution when a circular trig substitution worked as well. In my mind, I thought "well the answer is in terms of more elementary functions, and more people will understand this method", but now I think "hyperbolic expressions almost always are more elegant than their equivalent circular trig, the identities are so much more elegant and easier to remember as well as derive, and the integrals are almost always easier to solve".

I'm just pointing you what you can already see, at the beginning of this thread I was quite a newbie, still very new to the idea of partial fractions or integration by parts, and usually was not able to choose the best substitution. You can be the judge whether or not I have improved =]

And I've seen some more elegant, although somewhat simple, integrals, you may want to try this one:

\int^1_0 \frac{\ln (1+x)}{x} dx

 

[Count Iblis]

\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin\left(\pi p\right)}

 

[sutupidmath]

\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin\left(\pi p\right)}

Is p from naturals or from Z( integer)??

 

[coomast]

The integral \int_0^1\frac{ln(1+x)}{x}\cdot dx can be evaluated by considering the series expansion of the numerator, i.e. ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot \frac{x^k}{k}. Putting this into the integral gives after dividing it by x, switching the integral and the sum (permitted due to converging series): \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}x^{k-1}dx. This is after evaluation: \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}. This can be split up into the following two series \left(1+\frac{1}{3^2}+\frac{1}{5^2}+...\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\right).
These are respectively equal to \frac{\pi^2}{8} and \frac{\pi^2}{24}. The result is therefore I=\frac{\pi^2}{12}. The series are in my personal formula map :-), I think they were comming from a Fourier series of some kind, I don't exactly remember.

It was a nice exercise.

The other integral \int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{sin(p\pi)} seems to be wrong to me. Isn't the power supposed to be p-1[\tex]? Then it can be transformed into a beta function and via complex contour integration solved for the result shown. Using the power given as it is, will result in something different. This can be found in more advanced textbooks :-) All this under the assumption that I&#039;m not mistaken, it&#039;s been a while since I have done these kind of things. If I find the time I will try to work it out in detail.

 

[d_leet]

\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin\left(\pi p\right)}

If p is an integer, that is obviously false as sin(p*pi)=0.

 

[Count Iblis]

The integral converges for 0&lt;p&lt;1. Also, to address the point raised by coomast, changing p to p-1 would introduce a minus sign. Sometimes you see the numerator given as x^{p-1}. If we change this to x^{p} we get a minus sign and if we change p to minus p we get another minus sign. To check the sign, note that the integral is positive if it converges and so is the answer.

 

[mathwonk]

my favorite way to do odd powers of sec is by parts. it reduces the integral's power by 2 in each step. until finally you need to know just the integral of sec. (i hate double angle formulas in integrals as they make them seem complicated and tedious).

 

[PowerIso]

I agree with Mathwonk. I find integration by parts to be easier it comes to sec with odd powers. I'm pretty bad at remembering the trig tricks, so in the end, i'll goof it up. I like the safe route.