Which Integrals Are Fun and Special?

[Nebuchadnezza]

Here I have collected a large set of rather fun and somewhat special integrals. I have tried to order them from easiest to hardest. Please pm me if you:
1. think some of these integrals are not enjoyable.
2, If you have a fun integral.
3. If you think an integral is unsolvable.
4. If you think I should swap the ordering around.


The medium integrals and "OH MY GOD I NEED TO PLAY BANJO AND CRY" Integrals are coming up tomorrow. Although I think many of the veterans in here will have no problems with any of the integrals I am posting...


Please enjoy the first batch of the set:


Easy as 3.14159265

$$I \, = \, \int\limits_0^{\frac{\pi }{{3n}}} {\tan \left( {nx} \right) dx$$

$$I \, = \, \int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$$

$$I \, = \, \int\limits_1^{{e^n}} {\ln \left( x \right)} dx$$

$$I \, = \, \int {\frac{1}{{x\ln x}} dx$$

$$I \, = \, \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\sin {{\left( x \right)}^2} dx$$

$$I \, = \, \int\limits_0^{\ln \left( 2 \right)} {\ln \left( {x + 1} \right){e^x} dx$$

$$I \, = \, \int {\frac{x}{{{e^x}}} dx$$

$$I \, = \, \int\limits_0^\pi {x\left( {\sin \left( x \right) + \cos \left( x \right)} \right) dx$$

$$I \, = \, \int {\frac{{{e^x} + 1}}{{{e^x} - 1}}} dx$$

$$I \, = \, \int {{2^x}{e^x}} dx$$

$$I =\int\limits_0^{\frac{3}{4}\pi } {\sin \left( x \right)\cos \left( x \right) dx$$

$$I \, = \, \int {{e^{x + {e^x}}}} dx$$

$$I \, = \, \int {\frac{{{x^2} + 2x + 2}}{{x + 1}} dx$$

$$I \, = \, \int {\frac{{x + b}}{{x - c}} dx$$

$$I \, = \, \int\limits_0^\infty {{x^3}{e^{ - x}} dx$$

$$I \, = \, \int {\frac{{x + 1}}{{x - 1}} dx$$

$$I \, = \, \int {x{a^x}} dx$$

$$I \, = \, \int {{e^x}\sin \left( x \right)} dx$$

Does the following inequality hold?

$${I_{B}} > {I_{A}} + {I_{C}}$$

$$\text{Where}\;{I_{A}} = \int\limits_0^1 {\frac{1}{{1 + \sqrt x }} dx \; , \; {I_{B}} = \int\limits_0^1 {\frac{1}{{x + \sqrt x }}} dx \; \; \text{and}\;\;\,{I_{C}} = \int\limits_0^1 {\frac{{\sqrt x }}{{1 + \sqrt x }}} dx$$

$$I \, = \, \int\limits_0^4 {\frac{1}{{1 + \sqrt x }}} dx$$

$$I \, = \, \int {\sqrt {4 - x} } dx$$

$$\text{Is} \, 12 \,>\, \int\limits_0^3 {\sqrt x } dx \, ?$$

$$I \, = \, \int {\frac{1}{{x\ln {{\left( x \right)}^n}}} dx$$

Find the area between the function $$f(x)$$ and the x-axis when $$f(x)=\sqrt{a-\sqrt{x}}$$

$$I \, = \, \int\limits_0^4 {\frac{{\ln \left( x \right)}}{{\sqrt x }} dx$$

$$I \, = \, \int\limits_1^{{{\left( {\ln \left( a \right)} \right)}^2}} {{e^{\sqrt x }}} dx$$ and $$a \, \ge \, 1$$

$$I \, = \, \int\limits_{ - 1}^1 {\frac{{x + 1}}{{{{\left( {x + 2} \right)}^4}}} dx$$

$$\text{Show that}\int {\left( {x + 3} \right){{\left( {x - 1} \right)}^5} dx \text{is equal to} \frac{1}{{21}}\left( {3x + 11} \right){{\left( {x - 1} \right)}^6} + C$$

Medium

$$I \, = \, \int\limits_{\arccos \left( b \right)}^{\arcsin \left( b \right)} {\sin {{\left( x \right)}^2} dx$$

 

[Nebuchadnezza]

Here are all the medium challenges.
All of the following problems can be solved purely with methods learned in CALCII. Though these often require the use of more than one method and loads of tenacity. Especially in the later half of these problems. Some of these also have some special solutions... Well please enjoy yourself doing these problems. The fun integrals should be out soon too.

A new post because of for some odd reason I can not edit my old post. Otherwise I would of course have done this.

Medium rare

$$I \, = \, \int {\left( {1 + 2{x^2}} \right){e^{{x^2}}}} \, dx$$

$$I \, = \, \int {\sin \left( {\ln \left( x \right)} \right) \, dx$$

$$I \, = \, \int {\frac{{\cos \left( x \right) - \sin \left( x \right)}}{{\sin \left( x \right) + \cos \left( x \right)}}} \, dx$$

$$I \, = \, \int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} \, dx$$

Find the n`th integral of the following functions

$$1 = \sqrt x$$

$$2 = \frac{1}{x}$$

$$3 = \frac{{{e^x}}}{{{a^x}}}$$

$$I \, = \, \int {\frac{1}{{\sqrt {\left( {x - t} \right)\left( {x + t} \right)} }} \, dx$$

$$I \, = \, \int {\frac{{\sqrt {1 - {x^2}} }}{{{x^2}}}} \, dx$$

$$I \, = \, \int {\frac{{{2^x}{3^x}}}{{{9^x} - {4^x}}} \, dx$$

$$I \, = \, \int {\ln \left( {\frac{{2 - x}}{{2 + x}}} \right)} \, dx$$

$$I \, = \, \int\limits_{\sqrt 2 }^2 {\frac{1}{{{x^3}\sqrt {{x^2} - 1} }}} \, dx$$

$$I \, = \, \int {\frac{{1 + 2{x^2}}}{{{x^5}{{\left( {1 + {x^2}} \right)}^3}}} \, dx$$

$$I \, = \, \int\limits_{ - \infty }^\infty {\frac{{{x^2}}}{{{{\left( {{x^2} + {a^2}} \right)}^2}}} \, dx} \; \text{where} \; a\,>\, 0$$

$$\text{The integral} \; \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2} \, dx } \; \text{can be written as} \; {\left( {\frac{a}{b}} \right)^b \, \text{where a and b are integers. Find a + b}$$.

$$I \, = \, \int {\frac{{{{\left( {x + 2} \right)}^7}}}{{{{\left( {x + 7} \right)}^2}}} \, dx$$

$$I \, = \, \int {\frac{{\left( {1 + {x^2}} \right)}}{{\left( {1 - {x^2}} \right)\sqrt {1 + {x^4}} }} \, dx$$

$$I \, = \, \int {\sqrt {\frac{{k + x}}{x}} \, dx$$

$$I \, = \, \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \, dx$$

$$\text{Find the area restricted by the x-axis} \, , \, f(x)=x \, \text{and} \, g(x)=x^{\frac{1}{n}} \\ \; \text{What happens as n goes to infinity?}$$

$$I \, = \, \int\limits_{ - \pi }^\pi {\sqrt {1 + \cos \left( x \right)} } \, dx$$

$$I \, = \, \int\limits_0^{\frac{1}{2}} {\arcsin \left( {\sqrt x } \right)} \, dx$$

$$I \, = \, \int\limits_0^1 {\ln \left( {\sqrt {1 - x} + \sqrt {x + 1} } \right)} \, dx$$

$$I \, = \, \int\limits_0^2 {\sqrt {\frac{x}{{x - 2}}} \, dx$$

$$I \, = \, \int\limits_{\frac{\pi }{2}}^\pi {{e^{\frac{x}{2}}}\left( {\frac{{2 - \sin \left( x \right)}}{{1 - \cos \left( x \right)}}} \right)} \, dx$$

$$I \, = \, \int\limits_0^{\ln \left( 2 \right)} {\sqrt {\frac{{{e^x} + 1}}{{{e^x} - 1}}} } \, dx$$

$$I \, = \, \int {\sqrt {\tan \left( x \right)} } \, dx$$

Hopefully you see the correct images, some latex bug that keeps bugging me while editing

 

[Nebuchadnezza]

Here is the last set of problems, the "OH GOD I NEED TO CRY AND PLAY BANJO" set or simply named fun.

Fun

$$I \, = \, \int {\left( {1 + 2{x^2}} \right){e^{{x^2}}}} \, dx$$

$$I \, = \, \int\limits_0^\infty {\frac{{x - 1}}{{{x^5} - 1}}\, dx$$

$$I \, = \, \int\limits_0^\infty {\frac{1}{{{x^n} + 1}}} \, dx \,{\rm{ n}} \in \, R$$

$$I \, = \, \int\limits_0^\infty {\frac{{{x^{ - p}}}}{{1 + x}}\, dx = \frac{\pi }{{\sin \left( {\pi p} \right)}}}$$

$$I \, = \, \int\limits_0^\infty {\frac{{\sin \left( x \right)}}{x}\, dx$$

$$I \, = \, \int\limits_{ - 1}^1 {\frac{{\sin \left( \beta \right)}}{{1 - 2x\cos \left( \beta \right) + {x^2}}}\, dx$$

$$\int\limits_0^\infty {\frac{{\sin \left( {ax} \right)}}{{\sin \left( {bx} \right)}}\frac{1}{{1 + {x^2}}} = \frac{\pi }{2}\frac{{\sinh \left( a \right)}}{{\sinh \left( b \right)}}\;\text{if}\;\left| a \right|} < \left| b \right| \$$

$$I \, = \, \int\limits_0^\infty {\frac{{\ln {{\left( x \right)}^2}}}{{{x^2} + {a^2}}}\, dx$$

$$I \, = \, \int\limits_0^2 {\sqrt {{x^3} + 2 - 2\sqrt {1 + {x^3}} } + } \sqrt {{x^3} + 10 - 6\sqrt {1 + {x^3}} } \, dx$$

$$I \, = \, \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{1 + {x^2}}}\, dx$$

$$I \, = \, \int\limits_0^\infty {\left\lfloor x \right\rfloor } {e^{ - x}}\, dx$$

$$I \, = \, \int\limits_0^{\frac{\pi }{2}} {\tan^{-1}}\left( x \right) + {{\cot }^{ - 1}}\left( x \right)\, dx$$

$$I \, = \, \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{x}\, dx$$

$$I \, = \, \int\limits_0^\pi {\ln \left( {\sin x} \right)\, dx$$

$$I \, = \, \int {\frac{{{x^n}}}{{1 + x + \frac{{{x^2}}}{{2!}} + ... + \frac{{{x^n}}}{{n!}}}}\, dx {\rm{ }}\;where\;{\rm{ }}\;n\, >\, 0}$$

$$I \, = \, \int\limits_0^\pi {\frac{{x\sin x}}{{1 + {{\sin }^2}x}}} \, dx$$

$$I \, = \, \int\limits_0^1 {\frac{{{x^2}\ln \left( x \right)}}{{\sqrt {1 - {x^2}} }}\, dx$$

$$I \, = \, \int\limits_0^1 {\ln \left( x \right) \cdot \ln \left( {1 + x} \right)} \, dx$$

$$I \, = \, \int\limits_0^{\frac{\pi }{2}} {\frac{1}{{1 + \tan {{\left( x \right)}^\pi }}}} \, dx$$

$$I \, = \, \int\limits_0^\infty {\frac{{1 - \cos \left( x \right)}}{{{x^2}}}} \, dx$$

$$I \, = \, \int\limits_{ - \infty }^0 {\frac{{\sin \left( x \right) + \cos \left( x \right)}}{{{e^{ - x}} + \sin \left( x \right)}}\, dx$$

$$I \, = \, \int\limits_2^4 {\frac{{\sqrt {\ln \left( {9 - x} \right)} }}{{\sqrt {\ln \left( {3 + x} \right)} + \sqrt {\ln \left( {9 - x} \right)} }}\, dx$$

$$I \, = \, \int\limits_0^1 {\frac{1}{{\sqrt {\ln \left( {\frac{1}{x}} \right)} }}\, dx = \sqrt \pi }$$

$$I \, = \, \int\limits_0^\infty {{e^{ - \pi {x^2}}}} \, dx$$

$$I \, = \, \int\limits_0^\infty {\sin \left( {{x^2}} \right)\, dx$$

$$I \, = \, \int\limits_0^\pi {\frac{x}{{1 + \tan \left( x \right)}}} \, dx$$

$$I \, = \, \int\limits_0^\pi {\ln \left( {1 - 2a\cos x + {a^2}} \right)\, dx = 0\;\text{if}\;a \in \left( { - 1,1} \right) \; \text{else} \; 2\pi \ln \left| a \right|}$$

$$I \, = \, \int\limits_0^\infty {\frac{x}{{{e^x} + 1}}} \, dx$$

$$I \, = \, \int\limits_0^\infty {\frac{{{e^{ - 3x}} - {e^{ - 4x}}}}{x}} \, dx$$

$$I \, = \, \int\limits_0^{\frac{\pi }{2}} {\sqrt[3]{{\tan \left( x \right)}}\, dx$$

$$I \, = \, \int\limits_0^{\frac{1}{2}} {\arcsin \left( {\sqrt x } \right)\, dx$$

$$I \, = \, \int\limits_{ - \pi }^\pi {\frac{{\sin \left( {nx} \right)}}{{\left( {{2^x} + 1} \right)\sin \left( x \right)}}\, dx$$

$$I \, = \, \int\limits_0^1 {\frac{{\arctan \left( x \right)}}{{x + 1}}\, dx$$

$$I = \int\limits_0^1 {{x^m}\ln {{\left( x \right)}^n}dx = \frac{{n!{{\left( { - 1} \right)}^n}}}{{{{\left( {m + 1} \right)}^{n + 1}}}} + C{\rm{ \; where \; n}}{\rm{,m}} \in {\rm{N}}}$$

$$I = \int\limits_0^{\frac{\pi }{2}} {\cos {{\left( x \right)}^n}\cos \left( {nx} \right)dx = \frac{\pi }{{{2^{n + 1}}}}{\rm{ \; where \; n}}{\rm{,m}} \in {\rm{N}}}$$

$$I = \int\limits_0^\infty {\frac{1}{{{{\left( {x + \sqrt {1 + {x^2}} } \right)}^n}}} = \frac{n}{{{n^2} - 1}}}$$

$$I = \int\limits_0^\infty {{e^{ - ax}}{{\left( {{e^{ - x}} - 1} \right)}^n}} dx = {\left( { - 1} \right)^n}\frac{{n!\left( {a - 1} \right)!}}{{\left( {a + n} \right)!}}$$

$$I = \int_{-\infty}^{\infty} {\frac{1}{{{{\left( {{x^2} + 1} \right)}^k}}}dx} = \frac{{8\pi \left( {2k - 3} \right)!}}{{{4^k}\left( {k - 1} \right)!\left( {k - 2} \right)!}}$$

 

[homology]

Thanks for the challenges, if you have time it would be neat if these were TeXed up as a pdf that we could download, sort of like the crosswords...for math nerds.

I'm pretty decent at integration but I came across a neat method recently I figure is relevant: http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf

Apparently Feynman was pretty savvy at differentiating under the integral (I'm not sure of the historical accuracy). In any event this link shows how you can take an integral and make a similar integral with a new parameter that reduces to the old then the new parameter takes on some value. So you go ahead and differentiate the new integral with respect to the parameter and then integrate the, hopefully, easier integral. Its pretty neat.

Cheers,

Kevin

 

[Nebuchadnezza]

Thanks for the feedback =)

Will do so, but not until someone show some interest in this thread.

I spent a great deal of time looking through dozens of pages, and reading in old dusty books to find some of these challenges. So it would be great if anyone would take a stab at any of them.

The hard integrals indeed needs special methods, contour integration, differentiation under the integral sign, noticing symmetry, series expansion and so on is necessary to even begin solving these.

$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(b+a-x)dx$$

$$\int_{-a}^{a}f(x)dx=\int_{0}^{a}f(x)+f(-x)dx$$

I need to fix some of my easy integrals... Soon. But you know, you don't have to try to solve any of the hard ones, I find the easy and medium ones fun as well...

 

[snipez90]

0. Integrating by parts, one can show that
$$I \, = \, \int\limits_0^\infty {\frac{{\sin \left( x \right)}}{x}\, dx \, = \, \int\limits_0^\infty \frac{1 - \cos x}{x^2}\, dx$$
and the latter is slightly simpler to contour integrate due to the quadratic term in the denominator of the integrand.

1. $$I \, = \, \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{1 + {x^2}}}\, dx$$
Solution:

$I = \frac{\pi \ln 2}{8}$

Make the standard trig substitution x = tan(t) so that the integral becomes
$$I = \int_{0}^{\frac{\pi}{4}} \ln(\tan t + 1) \, dt = \int_{0}^{\frac{\pi}{4}} \ln\left(\frac{\sin t + \cos t}{\cos t}\right) \, dt$$

Since $\sin t + \cos t = \sqrt{2} \sin(x + \frac{\pi}{4})$ we have
$$I = \int_{0}^{\frac{\pi}{4}} \ln \sqrt{2} \, dx + \int_{0}^{\frac{\pi}{4}} \ln\sin\left(x+\frac{\pi}{4}\right) \, dx- \int_{0}^{\frac{\pi}{4}} \ln\cos x \, dx$$

The last two integrals cancel each other since
$$\int_{0}^{\frac{\pi}{4}} \ln\sin \left(x+\frac{\pi}{4}\right) \, dx = \int_{0}^{\frac{\pi}{4}} \ln \sin\left(\frac{\pi}{2} - x \right) \, dx = \int_{0}^{\frac{\pi}{4}} \ln\cos x \, dx$$

whence the conclusion follows.
2. $$I \, = \, \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{x}\, dx$$

Substituting u = ln(x+1) seems to reduce it to something manageable.

3. $$I \, = \, \int\limits_2^4 {\frac{{\sqrt {\ln \left( {9 - x} \right)} }}{{\sqrt {\ln \left( {3 + x} \right)} + \sqrt {\ln \left( {9 - x} \right)} }}\, dx$$

Apply $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(b+a-x)dx$ and add.

4. $$I \, = \, \int\limits_0^1 {\frac{1}{{\sqrt {\ln \left( {\frac{1}{x}} \right)} }}\, dx = \sqrt \pi }$$

The natural substitution u = ln(1/x) seems to work.

 

[Nebuchadnezza]

I have actually Tex up all of these... Many of them has been removed or changed into better ones. I also found allot more tough integrals. Will post the link later today

Thanks for the feedback! Cheers. Even with basic understanding of calculus you should be able to solve all of the easy problems.

Here is the tex file

http://www.2shared.com/document/AJV8VMSR/Integrals_from_R_to_Z.html

http://www.viewdocsonline.com/document/318q1d

If anyone has a direct hosting site that would be great.

8 pages. 1 page with just integrals 1 page with problems. For easy,medium and hard integrals.

I will probably name them, and add solutions when the complete list is up. For that I need you guys to tell which integrals are harder than one another. Like integral 2 is supposed to be a bit harder than integral 1 and so on. Also if you have any fun integrals missing feel free to post them here. Fun as in not necessary just an insane amount of computation.

Also please say if any of the integrals do not have a solution, or if a integral is not fun.

Took me forever gathering all of these, and making some up myself. Any feedback would be awesome.

 

[snipez90]

5. The ln(sin) one uses the same techniques as the the ln(x+1)/(1+x^2) one I did above. Just note sin(x) = (1/2)sin(x/2)cos(x/2) and keep exploiting symmetry to obtain cancellation.

6. $$I \, = \, \int\limits_0^\infty {\frac{{{e^{ - 3x}} - {e^{ - 4x}}}}{x}} \, dx$$

Let $\varepsilon > 0$. Split the integral and make the substitution u = 3x and subsequently u = 4x to obtain
$$\int_{\varepsilon}^{\infty} \frac{e^{-3x}-e^{-4x}}{x}\,dx = \int_{3\varepsilon}^{4\varepsilon} \frac{e^{-u}}{u}\, du$$
Letting $\varepsilon \rightarrow 0$, this last integral approaches
$$\int_{3\varepsilon}^{4\varepsilon}\frac{du}{u} = \log(4/3)$$.

More generally
$$\int\limits_0^\infty {\frac{{{e^{ - ax}} - {e^{ - bx}}}}{x}} \, dx = \log(b/a)$$
should hold provided Re(a), Re(b) are nonnegative if a and b are complex numbers.

Letting a = 1, b= i and setting the imaginary parts of the last equation equal, we get

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

7. Here is a direct approach to the sin x / x integral:
We work with $$\frac{\partial}{\partial \alpha}\int_{-\infty}^{\infty}\frac{\sin{\alpha\cdot x}}{x}\,dx$$ in a distributional sense.
Rewrite the integrand to introduce Fourier phases as follows
$$\frac{\partial}{\partial \alpha}\int_{-\infty}^{\infty}\frac{\sin{\alpha\cdot x}}{x}\,dx = \frac{\partial}{\partial \alpha}\int_{-\infty}^{\infty}\frac{e^{i\alpha x} - e^{-i\alpha x}}{2ix}\,dx.$$
Differentiating under the integral sign gives
$$\frac{\partial}{\partial \alpha}\int_{-\infty}^{\infty}\frac{\sin{\alpha\cdot x}}{x}\,dx = \int_{-\infty}^{\infty}\frac{\partial}{\partial \alpha}\frac{e^{i\alpha x} - e^{-i\alpha x}}{2ix}\,dx = \frac{1}{2}\int_{-\infty}^{\infty}\left(e^{i\alpha x} + e^{-i\alpha x}\right)dx.$$
The Fourier transform (non-unitary angular frequency convention) of the constant function f(x) = 1, is the Dirac delta multiplied by $2\pi$ whence
$$\frac{\partial}{\partial \alpha}\int_{-\infty}^{\infty}\frac{\sin{\alpha\cdot x}}{x}\,dx = 2\pi\delta(\alpha).$$
The Dirac delta is the distributional derivative of the Heaviside step function. Hence integrating both sides of the last equation with respect to $\alpha$ from -c to c shows that, in the sense of generalized functions, we have
$$\int_{-\infty}^{\infty}\frac{\sin{c\cdot x}}{x}\,dx = \pi\mbox{sgn}(c).$$
Here we need convergence in the ordinary sense of the left hand side for c = 1 (comes from the integration by parts argument I outlined in the last post). The same argument shows convergence in this sense for c in a neighborhood of 1. Integrating against test functions supported about c = 1 we conclude that the integral evaluates to $\pi$. Parity symmetry gives
$$\int_{0}^{\infty}\frac{\sin x}{x}\, dx = \frac{\pi}{2}$$

 

[Nebuchadnezza]

First of all, here is a fun challenge. Could anyone help me word it better in english?

"Let $$P(x)$$ be a third degree polynomial with two distinct roots. The integral between the roots is equal to $$\frac{27}{4}$$. Find $$P(x)$$"

And here is another way to solve this integral. $$I\left( x \right) = \int\limits_0^\infty {\frac{{\sin x}}{x}dx}$$

$$I\left( x \right) = \int\limits_0^\infty {\frac{{\sin x}}{x}{e^{ - bx}}dx}$$

If we make $$b=0$$ we obtain our original integral. Also note that the limits as x approaches both 0 and infinity are the same for the integral. Lastly we note that the integral is differentiable for all b`s. As long as b is positive.
This is kinda important to remember, we can't just wildly add a new constant, we have to check the limits, continuity and if the function is differentiable.

$$I'\left( x \right) = \frac{d}{{dx}}\int\limits_0^\infty {\frac{{\sin x}}{x} \cdot {e^{ - bx}}dx} = \int\limits_0^\infty {\frac{\partial }{{\partial x}}\frac{{\sin x}}{x} \cdot {e^{ - bx}}dx}$$

$$I'\left( x \right) = \int\limits_0^\infty {\sin x \cdot {e^{ - bx}}dx}$$

$$I'\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{{{e^{ - bx}}\left( {\cos \left( x \right) + \sin \left( x \right)b} \right)}}{{{b^2} + 1}}} \right]_0^n$$

$$I'\left( x \right) = - \frac{1}{{{b^2} + 1}}$$

$$I\left( x \right) = - \arctan \left( b \right) + C$$

To determine $$C$$ we let $$I\left( x \right) = 0$$ this happens as $$b$$ goes towards infinty

$$\mathop {\lim }\limits_{b \to \infty } \int\limits_0^\infty {\frac{{\sin x}}{x}{e^{ - bx}}dx} = \mathop {\lim }\limits_{b \to \infty } - \arctan \left( b \right) + C$$

$$0 = - \frac{\pi }{2} + C \Rightarrow C = \frac{\pi }{2}$$

$$\int\limits_0^\infty {\frac{{\sin x}}{x}{e^{ - bx}}dx} = - \arctan \left( b \right) + \frac{\pi }{2}$$

$$\int\limits_0^\infty {\frac{{\sin x}}{x}{e^{ - 0x}}dx} = - \arctan \left( 0 \right) + \frac{\pi }{2}$$

$$\underline{\underline {\int\limits_0^\infty {\frac{{\sin x}}{x}dx} = \frac{\pi }{2}}}$$

Also another one of your integrals could have been done much simpler with this substitution. Aka noting the symmetry $$I = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\tan x + 1} \right)dx}$$ here we exploit the fact that $$\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)}$$

$$I = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\tan \left( {\frac{\pi }{4} - x} \right) + 1} \right)dx} = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \tan x}}{{1 + \tan x}} + 1} \right)} = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\frac{2}{{1 + \tan x}}} \right)}$$

$$2I = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\tan x + 1} \right)dx} + \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\frac{2}{{1 + \tan x}}} \right)} dx$$

$$2I = \int\limits_0^{\frac{\pi }{4}} {\ln \left( 2 \right)} dx$$

$$I = \frac{{\pi \ln \left( 2 \right)}}{8}$$

$$\underline{\underline {I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\ln \left( {x + 1} \right)}}{{{x^2} + 1}}dx} = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\tan x + 1} \right)dx} = \frac{{\pi \ln \left( 2 \right)}}{8}}}$$

 

[Nebuchadnezza]

Yeah you have =) Only problem is that how to solve it might not be so clear... I have solved it though, and it was fun! Anyone dare to attempt an solution ?

 

[spamiam]

$$I\left( x \right) = \int\limits_0^\infty {\frac{{\sin x}}{x}dx}$$

$$I\left( x \right) = \int\limits_0^\infty {\frac{{\sin x}}{x}{e^{ - bx}}dx}$$

If we make $$b=0$$ we obtain our original integral. Also note that the limits as x approaches both 0 and infinity are the same for the integral. Lastly we note that the integral is differentiable for all b`s. As long as b is positive.
This is kinda important to remember, we can't just wildly add a new constant, we have to check the limits, continuity and if the function is differentiable.

$$I'\left( x \right) = \frac{d}{{dx}}\int\limits_0^\infty {\frac{{\sin x}}{x} \cdot {e^{ - bx}}dx} = \int\limits_0^\infty {\frac{\partial }{{\partial x}}\frac{{\sin x}}{x} \cdot {e^{ - bx}}dx}$$

$$I'\left( x \right) = \int\limits_0^\infty {\sin x \cdot {e^{ - bx}}dx}$$

I'm a bit confused...x is a dummy variable in this integral. I is actually a function of b, so shouldn't it be I(b)? Actually, looking at what you have for I', I think that's what you meant.

 

[Nebuchadnezza]

Ofcourse the remaining should be I(b)... sorry for the confusion...

Right so here is possibly the final pdf... 176 Problems ranging from easy to insane.

http://www.2shared.com/document/IHlhRmVk/Integrals_from_R_to_Z.html

Just for those who do not trust a download...

http://www.viewdocsonline.com/document/hidh0l

A. L. Bruce. The problem has a very simple solution, that is why I wanted to be able to state the problem clearly. Of course I can post a solution, but it is much more fun watching others attempt the problem. Hints can be given if needed.

 

[snipez90]

The second approach to the ln(1+x)/(1+x^2) isn't much simpler, though perhaps slightly more elegant since you bothered to remember trig identities involving tan :P.

For the following you need n to be > 1 :
$$I \, = \, \int\limits_0^\infty {\frac{1}{{{x^n} + 1}}} \, dx \,{\rm{ n}} \in \, R$$

For the next one you do need a condition on p, but since this condition is less obvious than in the previous case, I won't give it away.
$$I \, = \, \int\limits_0^\infty {\frac{{{x^{ - p}}}}{{1 + x}}\, dx = \frac{\pi }{{\sin \left( {\pi p} \right)}}}$$

The first one is a standard example of something easily handled by contour integration. If I do find a nicer way of doing the second (nicer than the complex-analytic way) I will post, but I suspect it will depend on whether you think series expansions are somehow nice.

 

[Nebuchadnezza]

I thought I remember to apply conditions to all of them. For the second one I guess its valid for all numbers except integers.

Standard contour integration... Where did you learn to do contour integration with variable residudes :p Anyho it is a nice challenge for anyone who has not seen it before.

As for what tanges go, I just remember the double formula. Not neccecarily the excact values for tangens.

Yeah, the point for those two is contour integration, if you do find another way please let me know.

 

[pwsnafu]

For the next one you do need a condition on p, but since this condition is less obvious than in the previous case, I won't give it away.
.

Isn't it $$-2 < Re [p] < -1$$? Or something like that. The lower bound gives convergence at x=0 the other x=infinity.

As an aside does anyone know how to do $$\int_0^1 \log\Gamma(x)dx=\log\sqrt{2\pi}$$?

 

[Nebuchadnezza]

More integrals! Perhaps the last bump... I think I have bumped the number of problems up to 170, ranging from easy to hard. All of them fun, Though I feel I need a few more easy fun integrals...

Integrals from R to Z

Any comments would be good. When this is complete I will add numbering and a really short solution.

 

[snipez90]

Isn't it $$-2 < Re [p] < -1$$? Or something like that. The lower bound gives convergence at x=0 the other x=infinity.

As an aside does anyone know how to do $$\int_0^1 \log\Gamma(x)dx=\log\sqrt{2\pi}$$?

By the same symmetry exploit

$$I = \int_{0}^{1} \log \Gamma (1-x) \,dx \Rightarrow$$

$$2I = \int_{0}^{1} \log \Gamma (x) \cdot \Gamma (1-x) \,dx$$

By http://planetmath.org/encyclopedia/EulerReflectionFormula.html" this becomes

$$2I = \int_{0}^{1} \log \pi \,dx - \int_{0}^{1} \log \sin \pi x \, dx$$

You should check that the second integral evaluates to -log(2) (make a change of variables and it's basically one of the integrals in this thread; I outlined the idea to that integral a few threads back). The desired conclusion follows immediately.

 

[Nebuchadnezza]

Hahhahaha snipez90 seems like no matter what kind of integrals or integration problems I throw at you, I never manage to stump you. Awesome job.

Hopefuly some of these problems were fun =)

I am sad to see that the art of integration is slowly losing its pride due to computerpower, which I find sad...

If you want I could post my solution to

$$\int_{0}^{\frac{\pi}{2}}\ln(\sin(x))\,dx$$

I think this is the function snipes is hinting at.

 

[pwsnafu]

Yeah, the point for those two is contour integration, if you do find another way please let me know.

We know that the Laplace transform of the constant is 1/s. Let s = 1+x. Then

$$\int_0^\infty \frac{x^{-p}}{1+x}dx = \int_0^\infty x^{-p} \left( \int_0^\infty e^{-t}e^{-xt}dt\right)dx$$

Change the order of integration and evaluating the Laplace we have

$$\int_0^\infty e^{-t}\frac{\Gamma(1-p)}{t^{1-p}}dt = \Gamma(1-p)\Gamma(p)$$

 

[snipez90]

Hahhahaha snipez90 seems like no matter what kind of integrals or integration problems I throw at you, I never manage to stump you. Awesome job.

Hopefuly some of these problems were fun =)

Haha you give me too much credit. The truth is I was probably in a similar position towards the end of high school as you are now, so I have seen quite a few of these before. I know that the ln(sin) one is well-known and I think appeared on the Putnam in the 1950's. Fortunately I was able to solve that on my own awhile back just thinking about sin, cos symmetry. I did not solve the ln(1+x)/(1+x^2) on my own. For me the trickiest step the solution I gave is rewriting sin + cos. It was the only step that I had to ask about back then and the only step that gave me some pause now. Others such as sinx/x I've seen many ways of doing, though I do like my solution, which requires a good deal of (real) analysis to fully justify.

But there are many that I haven't done. For instance the ln(x)*ln(1+x) seems pretty tough, and the most amenable approach I've thought of is series expansion (expand ln(1+x)).

Props for consolidating the integrals. I'll revisit them as I get around to reviewing the basic techniques.

Also I made an error in my last post. The very last integral should evaluate to -log(2).

 

[Nebuchadnezza]

You could also take a look at the easy and medium integrals. They sort of force you to refine your basic techniques. I was able to steal some of Eulers works, many nice integrals in that book!

 

[Nebuchadnezza]

A smal bump and a few more fun integrals ^^

$$I_1 \, = \, \int{x^x \( \ln(x)\,+\,1)}\,dx$$

$$I_3 \, = \, \int_{-\infty}^{\infty}x^a\cdot e^{-b} \, dx$$

$$I_2 \, = \, \int{\sqrt{\,x^2 + 5x + 13\,}} \, dx$$

$$I_5 \, = \, \int_{-1}^{1}{\arctan{(e^x)} \, dx}$$

 

[Nebuchadnezza]

A smal bump, have updated the document with even more integrals. Any comments at all? ...

http://www.2shared.com/document/Djbt3aIn/Integrals_from_R_to_Z.html

I tried my best to only pick fun and somewhat special integrals... Judging by the response here, seems nobody liked them.