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PostPosted: Sat, 23 Feb 2008 20:06:11 UTC 
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Find :

$ I= \int _ {\phi=0} ^{ \frac {\pi } 2 } \int _ { \theta = 0 } ^ { \frac {\pi} 2 } \frac { \ln ( 2- \sin \theta \cos \phi ) \sin \theta }{ 2- 2 \sin \theta \cos \phi + \sin ^2 \theta \cos ^2 \theta }  d\theta \ d\phi


Last edited by M.A on Fri, 14 Mar 2008 19:16:57 UTC, edited 3 times in total.

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PostPosted: Sun, 24 Feb 2008 05:58:39 UTC 
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¬logical wrote:
dragade3789 wrote:
Image

Are the denominators of the 2nd and 3rd expression purposely equal or is it a mistake?


i think that is on purpose.


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PostPosted: Sun, 24 Feb 2008 06:00:10 UTC 
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dragade3789 wrote:
¬logical wrote:
dragade3789 wrote:
Image

Are the denominators of the 2nd and 3rd expression purposely equal or is it a mistake?


i think that is on purpose.


Why? Usually such things have an interesting sort of symmetry for a reason.

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PostPosted: Mon, 25 Feb 2008 14:01:51 UTC 
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Prove that \displaystyle\int_0^1 (\ln x)^n dx=(-1)^n n! for n=1,2,3,...

Spoiler:
This one is quite simple.
Let \displaystyle I_n=\int_0^1(\ln x)^n dx, then integrate by parts.
Let \displaystyle u=(\ln x)^n, and dv=dx\Rightarrow v=x.
Therefore, \displaystyle I_n=[x(\ln x)^n ]^1_0 -\int_0^1 x\cdot n(\ln x)^{n-1}\cdot\frac{1}{x} dx
We obtain \displaystyle I_n=-n\cdot I_{n-1} ... etc.

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PostPosted: Tue, 4 Mar 2008 22:00:23 UTC 
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Prove that \displaystyle\int_0^{\infty}\frac{\tan^{-1}(\pi x)-\tan^{-1}(x)}{x}dx=\frac{\pi}{2}\ln\pi.

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PostPosted: Tue, 4 Mar 2008 22:50:04 UTC 
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bidentate wrote:
Prove that \displaystyle\int_0^{\infty}\frac{\tan^{-1}(\pi x)-\tan^{-1}(x)}{x}dx=\frac{\pi}{2}\ln\pi.


your integral = $ \int_0^{\infty} \int_1^{\pi} \frac{1}{1+x^2y^2}dydx=\int_1^{\pi} \int_0^{\infty}\frac{1}{1+x^2y^2}dxdy = \int_1^{\pi}\frac{\pi}{2y}dy=\frac{\pi}{2}\ln \pi.


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PostPosted: Wed, 2 Apr 2008 23:33:59 UTC 
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1) Using the auxiliary function $ f(x)=\left(\int_0^x e^{-t^2} dt \right)^2 + \int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1} dt, prove that $ \int_0^{\infty}e^{-t^2} dt =\frac{\sqrt{\pi}}{2}.

2) Fresnel Integrals: Using the complex auxiliary function $ g(x)=\left(\int_0^x e^{it^2} dt \right)^2 + i\int_0^1 \frac{e^{ix^2(t^2+1)}}{t^2+1} dt, prove that:

$ \int_0^{\infty} \sin(t^2) dt = \int_0^{\infty} \cos(t^2) dt = \sqrt{\frac{\pi}{8}}.


Hint:
Spoiler:
Find f'(x) and g'(x) first!


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PostPosted: Thu, 15 May 2008 00:39:33 UTC 
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Evaluate the limit: \displaystyle L=\lim_{y\to 0}\frac{1}{y}\int_0^{\pi}\tan(y\sin x)dx.

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PostPosted: Mon, 19 May 2008 16:04:15 UTC 
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bidentate wrote:
Evaluate the limit: \displaystyle L=\lim_{y\to 0}\frac{1}{y}\int_0^{\pi}\tan(y\sin x)dx.


Spoiler:
\displaystyle L=\lim_{y\to 0}\frac{1}{y}\int_0^{\pi}\tan(y\sin x)dx

=\displaystyle \int_0^{\pi} \lim_{y\to 0}\frac{\tan(y\sin x)-\tan(0 \sin x)}{y}dx

=\displaystyle \int_0^{\pi} \frac{\partial}{\partial y}(\tan(y\sin x))dx
evaluated at y = 0.

From here it is simple to show that the answer is 2.


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PostPosted: Mon, 19 May 2008 20:40:44 UTC 
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Zathras wrote:
Spoiler:
\displaystyle L=\lim_{y\to 0}\frac{1}{y}\int_0^{\pi}\tan(y\sin x)dx

=\displaystyle \int_0^{\pi} \lim_{y\to 0}\frac{\tan(y\sin x)-\tan(0 \sin x)}{y}dx

=\displaystyle \int_0^{\pi} \frac{\partial}{\partial y}(\tan(y\sin x))dx
evaluated at y = 0.

From here it is simple to show that the answer is 2.

Correct!

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PostPosted: Tue, 22 Jul 2008 20:21:50 UTC 
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Hi.

commutative wrote:
1) Using the auxiliary function $ f(x)=\left(\int_0^x e^{-t^2} dt \right)^2 + \int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1} dt, prove that $ \int_0^{\infty}e^{-t^2} dt =\frac{\sqrt{\pi}}{2}.

This one is a great way to prove the result ! ^^

Spoiler:
We first have to prove that the 2 integrals in f(x) (uniformly if I remember) converge. This part will be left to theoricists who like dominating :D

$f'(x)=2 e^{-x^2} \cdot \int_0^x e^{-t^2} \ dt-2x \int_0^1 e^{-x^2(t^2+1)} \ dt=2e^{-x^2} \left(\int_0^x e^{-t^2} \ dt-x \int_0^1 e^{-(xt)^2} \ dt \right)

Change variable of the second : $xt=u.

$f'(x)=2e^{-x^2} \left(\int_0^x e^{-t^2} \ dt-\int_0^x e^{-u^2} \ du\right)=0

Thus the function f is constant.
We can write in particular $f(0)=\lim_{x \to \infty} f(x)

$f(0)=0^2+\int_0^1 \frac{1}{t^2+1} \ dt=\arctan(t) \bigg|_0^1=\tfrac \pi 4

Therefore $\tfrac \pi 4=\lim_{x \to \infty} f(x)=\left(\int_0^\infty e^{-t^2} \ dt \right)^2 + \lim_{x \to \infty} \int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1} \ dt
Because this latter integral uniformly converges, we can invert the integral and the limit, which is 0.

$\therefore \quad \tfrac \pi 4=\left(\int_0^\infty e^{-t^2} \ dt \right)^2

$e^{-t^2} is a positive function, for any t. So its integral over $(0, \infty) is positive.

We can conclude that $\int_0^\infty e^{-t^2} \ dt=\frac{\sqrt{\pi}}{2}

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PostPosted: Wed, 15 Oct 2008 23:57:18 UTC 
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Here's a hard one... it was asked 10 yrs ago on a exam.

I can't even evaluate the borders becuz it gives an undetermined form and they'res a function in the result that seems very hard to derivate.

Image
Image


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