Unlock the Most Challenging Integrals with Jacob

[pierce15]

Hello all,

I started a thread exactly like this a few months ago. I really enjoy doing integrals, so if you could post some of the most challenging ones you know of, I would greatly appreciate it

Jacob

 

[Office_Shredder]

This isn't going to be a worldbeater in this thread but it's nontrivial

$$\int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx$$

 

[pierce15]

This isn't going to be a worldbeater in this thread but it's nontrivial

$$\int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx$$

... I'm not sure where to start with this one. I'm thinking integration by parts, but I'm yet to choose a u and dv that actually simplify the problem

 

[iRaid]

$$\int \frac{cos(x)}{1+sin^{3}(x)}dx$$

Have fun.

 

[Office_Shredder]

... I'm not sure where to start with this one. I'm thinking integration by parts, but I'm yet to choose a u and dv that actually simplify the problem

Oops, maybe it was too hard. What mathematical experience do you have? The typical solution to the integral (AFAIK) requires doing something that isn't normally taught in a calculus 1 or 2 course, although if you are really clever you can solve it using only techniques that my calculus 2 students have learned (but I would bet hundreds of dollars on none of them ever being able to solve it if I gave them a week and they didn't cheat)

 

[pwsnafu]

My favourite
$\int_{\pi/4}^{\pi/2} \log \log \tan x \, dx = \frac\pi2 \log\left(\frac{\sqrt{2\pi}\Gamma(3/4)}{\Gamma(1/4}\right)$

There's also

$\int_0^1 \log\left( \frac{\Gamma(x+t)}{\sqrt{2\pi}}\right) dx = t \log t -t$
for $t\geq 0$

 

[pierce15]

$$\int \frac{cos(x)}{1+sin^{3}(x)}dx$$

Have fun.

$$ u = \sin x, \quad du = \cos x $$

$$ = \int \frac{du}{1+u^3} = \int \frac{du}{(1+u)(1-x+u^2)} $$

At this point, I did a partial fraction decomp, rewriting the integral as:

$$ \int \left[ \frac{1/3}{1+u} + \frac{ -\frac{1}{3} u + 2/3}{1-u+u^2} \right] \, du $$

$$ =\frac{1}{3} \ln |1+u| + \int \left[ \frac{2/3}{1-u+u^2} - \frac{ \frac{1}{3} u }{1-u+u^2} \right] \, du $$

Doing those two integrals separately:

$$ \int \frac{2/3}{1-u+u^2} \, du $$

$$ \int \frac{2/3}{(u-1/2)^2 + 3/4} \, du = \frac{8}{9} \arctan \left( \frac{4}{3}(u-1/2) \right) $$

For the second one:

$$ \int \frac{ \frac{1}{3} u }{(u-1/2)^2 + 3/4} \, du $$

$$ z = u + 1/2 $$

$$ = \int \left[ \frac{ \frac{1}{3} z }{z^2 + 3/4} - \frac{1/6}{z^2 + 3/4} \right] \, dz $$

$$ = \frac{1}{6} \ln | z^2 + 3/4 | - \frac{2}{9} \arctan \left( \frac{4}{3} z \right) $$

Thus, the integral is:

$$ \int \frac{cos(x)}{1+sin^{3}(x)} \, dx = \frac{1}{3} \ln |1+ u| + \frac{8}{9} \arctan \left( \frac{4}{3}(u-1/2) \right) - \left[ \frac{1}{6} \ln |z^2 +3/4| - \frac{2}{9} \arctan \left( \frac{4}{3} z \right) \right] + C $$

$$ = \frac{1}{3} \ln |1+ \sin x| + \frac{10}{9} \arctan \left( \frac{4}{3}(u-1/2) \right) - \frac{1}{6} \ln |(u+1/2)^2 +3/4| + C $$

Thus, in conclusion,

$$ \int \frac{\cos(x)}{1+\sin^{3}(x)} \, dx = \frac{1}{3} \ln |1+ \sin x| + \frac{10}{9} \arctan \left( \frac{4}{3}\sin x-\frac{2}{3} \right) - \frac{1}{6} \ln \left| \left(\sin x+\frac{1}{2} \right)^2 +\frac{3}{4}\right| + C $$

 

[pierce15]

Oops, maybe it was too hard. What mathematical experience do you have? The typical solution to the integral (AFAIK) requires doing something that isn't normally taught in a calculus 1 or 2 course, although if you are really clever you can solve it using only techniques that my calculus 2 students have learned (but I would bet hundreds of dollars on none of them ever being able to solve it if I gave them a week and they didn't cheat)

I only know all of the techniques of integration taught in single variable calculus textbooks, as well as a few other tricks. Does this one involve using complex polar coordinates? I've seen integrals like that before, but I've never been clever enough to solve one myself.

 

[Office_Shredder]

I only know all of the techniques of integration taught in single variable calculus textbooks, as well as a few other tricks. Does this one involve using complex polar coordinates? I've seen integrals like that before, but I've never been clever enough to solve one myself.

You can do it in complex coordinates using the residue theorem (which is how I originally solved it). But you can also solve it in a nifty different way. Let
$$f(a) = \int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx$$

What is f'(a)?

 

[pierce15]

My favourite
$$\int_{\pi/4}^{\pi/2} \log \log \tan x \, dx = \frac\pi2 \log\left(\frac{\sqrt{2\pi}\Gamma(3/4)}{\Gamma(1/4}\right)$$

There's also

$$ \int_0^1 \log\left( \frac{\Gamma(x+t)}{\sqrt{2\pi}}\right) dx = t \log t -t$$
for $t\geq 0$

Those look pretty interesting... unfortunately, I don't think that I know enough about the gamma function to solve these

 

[pierce15]

You can do it in complex coordinates using the residue theorem (which is how I originally solved it). But you can also solve it in a nifty different way. Let
$$f(a) = \int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx$$

What is f'(a)?

I don't see how you can take the derivative of that function. Are you suposed to employ partial derivatives? If so, I don't know enough about calculus of several variables to do so

 

[Mandelbroth]

I don't see how you can take the derivative of that function. Are you suposed to employ partial derivatives? If so, I don't know enough about calculus of several variables to do so

Differentiation under the integral sign. I don't know when they teach it, but I'm pretty sure you know a special case of it. So, let's see if I can help you understand.

What is the relation between antiderivatives and derivatives?

 

[pierce15]

$$ \frac{d}{dx} \int_a^x f'(t) \, dt = f(x) $$

 

[pierce15]

Upon reading the Wikipedia article about Differentiation under the integral sign, here is what I've collected:

If

$$ f(z) = \int_a^b g(x,z) \, dx $$

(where a and b are constants), then

$$f'(z) = \int_a^b \frac{ \partial}{\partial z} g(x,z) \, dx $$

Is that correct?If so, I'll try to do the problem.

$$ f(a) = \int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx $$

$$ f'(a) = \int_0^\infty -x\cdot e^{-ax} \frac{\sin(x)}{x} \, dx $$

$$ = - \int_0^\infty e^{-ax} \cdot \sin(x) \, dx $$

 

[CompuChip]

Looks like a good candidate for partial integration :)

 

[Mandelbroth]

$$ \frac{d}{dx} \int_a^x f'(t) \, dt = f(x) $$

Close, but not quite. $\displaystyle \frac{d}{dx} \int_a^x f'(t) \, dt = \frac{d}{dx}[f(x)-f(a)] = f'(x)$. Though, the relationship I was looking for was $\displaystyle \frac{d}{dx}\int_a^b f(x,t) \, dt = \int_a^b \frac{\partial f}{\partial x} \, dt$.

Edit: piercebeatz, your answer to the derivative of the function in the integration problem is correct.

 

[pierce15]

Ok, I've got the derivative... but what now?

 

[Mandelbroth]

Ok, I've got the derivative... but what now?

My pre-calculus teacher likes to say, "If you don't know what to do, do something."

What do you get for the value of the differentiated integral? Your final answer should be fairly obvious to you at that point if you know your integral identities.

 

[pierce15]

Oh, I think I see what to do. I won't write out all my complete integration by parts substitutions for the sake of time.

$$ f'(a) = - \int_0^\infty e^{-ax} \cdot \sin x \, dx $$

$$u = e^{-ax}, \quad dv = - \sin x \, dx $$

$$ = e^{-ax} \cdot \cos x \big|_{x=0}^{x=\infty} + a \int_0^\infty e^{-ax} \cdot \cos x \, dx $$

$$ = 1 + a \int_0^\infty e^{-ax} \cdot \cos x \, dx $$

$$ u = e^{-ax}, \quad dv = \cos x \, dx $$

$$ = 1 + a \left[ e^{-ax} \cdot \sin x \big|_{x=0}^{x=\infty} +a \int_0^\infty e^{-ax} \cdot \sin x \, dx \right] $$

$$ f'(a) = - \int_0^\infty e^{-ax} \cdot \sin x \, dx = 1 + a^2 \int_0^\infty e^{-ax} \cdot \sin x \, dx $$

$$ f'(a) = (-1-a^2) \int_0^\infty e^{-ax} \cdot \sin x \, dx = 1 $$

$$ f'(a) = - \frac{1}{1+a^2} + C $$

$$ f(a) = - \arctan a + C $$

Going back to the original integral, we have:

$$ f(0) = \int_0^\infty \frac{\sin x }{x} \, dx $$

This will be the value of $C$... But how can I compute this?

 

[Mandelbroth]

$$ f'(a) = (-1-a^2) \int_0^\infty e^{-ax} \cdot \sin x \, dx = 1 $$

[...]

$$ f(0) = \int_0^\infty \frac{\sin x }{x} \, dx $$

This will be the value of $C$... But how can I compute this?

You need to learn what an equal sign means, because $f'(a)\neq 1$. However, your overall answer is correct.

I think I'd just take the value of f(0) to be a special case of the Borwein integrals. It's $\pi/2$.

So, your final answer is $-\arctan(a)+\frac{\pi}{2}$. In the original integral, $a$ must be greater than or equal to 0, so what does this simplify to?

 

[pierce15]

$$ - \arctan a + \frac{\pi}{2} = x $$

$$ \frac{ \tan (-\arctan a) + \tan \pi/2}{1 - \tan (- \arctan a) \cdot \tan \pi/2} = \tan x $$

$$ \frac{ 1-a}{1+a} = \tan x $$

$$ \int_0^\infty e^{-ax} \frac{ \sin x}{x} \, dx = \arctan \left( \frac{1-a}{1+a} \right) $$

How would one show:

$$ \int_0^\infty \frac{ \sin x}{x} \, dx = \frac{ \pi}{2} $$

?

 

[Bipolarity]

Isn't this problem motivated by Laplace transforms?

BiP

 

[Office_Shredder]

This problem was motivated by a Fourier transform originally. Sorry I missed a lot of action...
a=0 isn't the best place to look. What happens as a goes to infinity? You should be able to prove that f goes to zero directly (bound the integrand by e^-ax) and this gets you C directly

 

[Mandelbroth]

$$ \frac{ \tan (-\arctan a) + \tan \pi/2}{1 - \tan (- \arctan a) \cdot \tan \pi/2} = \tan x $$

$$ \int_0^\infty \frac{ \sin x}{x} \, dx = \frac{ \pi}{2} $$

?

$\tan\frac{\pi}{2}$ is undefined. I was looking for arccot(a).

How would you show the integral identity? It might be a little advanced for you, I think, but are you familiar with the Laplace transform?

The Laplace transform of a function f of t, denoted $\mathcal{L}\left\{f(t)\right\}$, is given by $$\mathcal{L}\left\{f(t)\right\}(s) = \int_0^\infty e^{-st}f(t)dt$$
for $s\in\mathbb{C}$.

With this in mind, it is relatively simple (though in no sense completely trivial) to show that
$$\int_0^\infty \frac{\sin x}{x} \, dx = \int_0^\infty \mathcal{L}\left\{sin(x)\right\}(s) \, ds = \int_0^\infty \frac{ds}{s^2+1} = \lim_{\alpha\rightarrow\infty}\arctan(\alpha) - \arctan(0) = \frac{\pi}{2}$$

 

[arildno]

To simplify Your answer, remember that arctan(x)+arctan(1/x)=pi/2

 

[mathnerd15]

$$\oint E\cdot dA=|E|\int_{0}^{2\pi}\int_{0}^{\pi}(1+1/2sin6\theta\sin5\phi)^2sin\phi d\phi d\theta =|E|\int_{0}^{2\pi}(\frac{25}{99}sin^2(6\theta)+2) d\theta =|E|\frac{421\pi}{99}=$$
$$\frac{\rho_{q}}{\varepsilon o}\int_{0}^{2\pi }\int_{0}^{\pi }\int_{0}^{(1+\frac{1}{2}sin6\theta sin5\phi )}\rho^2sin(\phi )d\rho d\phi d\theta=$$
$$\frac{\rho_{q}}{\varepsilon o} \int_{0}^{2\pi }\int_{0}^{\pi } \frac{1}{198}(157-25cos12\theta )d\phi d\theta= \frac{157\pi\rho_{q}}{99\varepsilon o }... E=\rho_{q}\frac{157}{421\varepsilon o}...Er_{1}-Er_{2} = \frac{\rho_{q}d157}{421\varepsilon o}$$

 

[MadAtom]

Am I actualy going to apply those hard integral in physics problems?

 

[pierce15]

Mathnerd15... what the hell is that?

 

[mathnerd15]

it is a variation on the electrical field between 2 wrinkled spheres of charge

 

[MadAtom]

Am I actualy going to apply those hard integral in physics problems?

It wasn't a rhetoric question. I'm really curious about it. Not being an ace at integrattion could, somehow, stop me (or slow me down) from solving a physics problem in the more advanced courses or even in my late career?