# Hard Integrals

1. May 9, 2013

### 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

2. May 9, 2013

### Office_Shredder

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

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

3. May 9, 2013

### pierce15

... 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

4. May 9, 2013

### iRaid

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

Have fun.

5. May 9, 2013

### Office_Shredder

Staff Emeritus
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)

6. May 10, 2013

### 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 $t\geq 0$

Last edited: May 10, 2013
7. May 11, 2013

### pierce15

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

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

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,

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

8. May 11, 2013

### pierce15

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.

9. May 11, 2013

### Office_Shredder

Staff Emeritus
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)?

10. May 11, 2013

### pierce15

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

11. May 11, 2013

### pierce15

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

12. May 12, 2013

### Mandelbroth

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?

13. May 12, 2013

### pierce15

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

14. May 12, 2013

### 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.

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

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

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

Last edited: May 12, 2013
15. May 12, 2013

### CompuChip

Looks like a good candidate for partial integration :)

16. May 12, 2013

### Mandelbroth

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.

17. May 12, 2013

### pierce15

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

18. May 12, 2013

### Mandelbroth

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.

Last edited: May 12, 2013
19. May 12, 2013

### 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?

20. May 12, 2013

### Mandelbroth

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?