1. Integration

Could someone post on here a semi-difficult integral if they wouldn't mind

2. Originally Posted by Mathstud28
Could someone post on here a semi-difficult integral if they wouldn't mind
$\displaystyle \int_\frac{-\pi}{4}^\frac{\pi}{4} \arccos(\tan x)\;dx$

i don't know if that's even remotely difficult for you. it was semi-difficult for me.

3. Originally Posted by polymerase
$\displaystyle \int \arccos(\tan x)\;dx$

i don't know if that's even remotely difficult for you. it was semi-difficult for me.
This integral does not have a elementary solution? You know that right? But I greatly extremely appreciate your effort!

4. Originally Posted by Mathstud28
This integral does not have a elementary solution? You know that right? But I greatly extremely appreciate your effort!
try it out....i forgot the intervals

5. Calculate $\displaystyle \int_0^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ bearing in mind that $\displaystyle \int_0^{\infty}{\frac{\sin(x)}{x}dx}=\frac{\pi}{2}$

It's not difficult but I still like the result

6. Originally Posted by polymerase
try it out....i forgot the intervals
Oh...that makes a big difference

Well lets see here $\displaystyle \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}arcos(tan(x))dx=\frac{\pi{ x}}{2}\bigg|_{\frac{-\pi}{4}}^{\frac{\pi}{4}}-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}arcsin(tan(x))dx$

Simplifying and adding the integrals we get an indentity $\displaystyle \frac{{\pi}^2}{4}=\frac{{\pi}^2}{4}$

so the answer is $\displaystyle \frac{{\pi}^2}{4}$

How did you do it? Modifying the phase shift?

7. Originally Posted by PaulRS
Calculate $\displaystyle \int_0^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ bearing in mind that $\displaystyle \int_0^{\infty}{\frac{\sin(x)}{x}dx}=\frac{\pi}{2}$

It's not difficult but I still like the result
Hmm...I know the way you are thinking of but I will try this

$\displaystyle \int_0^{\infty}\frac{\sin^2(x)}{x^2}dx=\frac{1}{2} \int_0^{\infty}\frac{1-\cos(2x)}{x^2}dx$

seperating we get $\displaystyle \frac{1}{2}\bigg[\int_0^{\infty}\frac{1}{x^2}dx-\int_0^{\infty}\frac{\cos(2x)}{x^2}dx\bigg]$

Now I dont feel like writing it all but I used power series and got $\displaystyle \frac{\pi}{2}$ as the answer as well

8. Originally Posted by PaulRS
Are you sure?

About the integral I posted: You should make use of some new techniques you've seen in this forum not long ago
Two things...what is wrong about that...that is the answer isnt it? I even checked my calculator once I was done ...and please dont tell me I have to use the imaginary part of $\displaystyle e^{-x(a-bi)}$ I am only seventeen and a junior in highschool...dont give me this complex analysis stuff lol

9. Define $\displaystyle J(p)=\int_{0}^{\infty}{\frac{\sin^2(p\cdot{x})}{x^ 2}dx}$ ...

...and now differentiate under the integral sign

10. Originally Posted by PaulRS
Define $\displaystyle J(p)=\int_{0}^{\infty}{\frac{\sin^2(p\cdot{x})}{x^ 2}dx}$ ...

...and now differentiate under the integral sign
Haha what part about junior in highschool, seventeen, does not know complex analysis...uhm...I will try....lets see here we want...Give me a little more hint...is this something to do with $\displaystyle Im[e^{ix}]$?

11. Originally Posted by Mathstud28
Haha what part about junior in highschool, seventeen, does not know complex analysis...uhm...I will try....lets see here we want...Give me a little more hint...is this something to do with $\displaystyle Im[e^{ix}]$?
Don't worry

Leibniz's integral rule

Differentiation under the integral sign - Wikipedia, the free encyclopedia

You do not need to use complex numbers

12. Originally Posted by PaulRS
Ok so if $\displaystyle J(p)=\int{\frac{\sin^2(px)}{x^2}}dx$

then $\displaystyle J'(p)=2\int{\frac{\sin(ax)\cos(ax)}{x}}dx$?
That is what I am getting from that...I took the partial in respect to a?

13. Here is an example of Leibniz's Rule in use

http://www.mathhelpforum.com/math-he...77-post19.html

And here is the complete thread: http://www.mathhelpforum.com/math-he...tegrals-2.html

14. Originally Posted by PaulRS
Here is an example of Leibniz's Rule in use

http://www.mathhelpforum.com/math-he...77-post19.html

And here is the complete thread: http://www.mathhelpforum.com/math-he...tegrals-2.html
So is what I did completely wrong?

15. Originally Posted by Mathstud28
Ok so if $\displaystyle J(p)=\int{\frac{\sin^2(px)}{x^2}}dx$

then $\displaystyle J'(p)=2\int{\frac{\sin(ax)\cos(ax)}{x}}dx$?
That is what I am getting from that...I took the partial in respect to a?
It should have been: $\displaystyle J'(p)=\int_0^{\infty}{\frac{2\sin(px)\cos(px)}{x}} dx$

Now relate it to $\displaystyle \int_0^{\infty}{\frac{\sin(x)}{x}}dx$

And finally integrate: $\displaystyle \int_0^tJ'(p)dp=J(t)-J(0)=J(t)$

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