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December 7th, 2011, 01:34 PM   #11
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Re: Nine Integrals

Quote:
 Originally Posted by guynamedluis Wow, Thanks a lot ZardoZ. Your proof of 5 is pretty slick.
[color=#000000]Regarding 5 and 8, it is clear that complex analysis does not always offer simple ways to compute integrals (wnvl's way is better of course). I believe that it was given to you for a better understanding of the techniques of contour integration![/color]

 December 29th, 2011, 06:38 PM #12 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals I'd like to contribute with 5. $\int\limits_{ - \infty }^\infty {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{n + 1}}}}}= 2\int\limits_0^\infty {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{n + 1}}}}}$ $x= \tan \phi$ $dx= {\sec ^2}\phi d\phi$ $2\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\phi d\phi }}{{{{\sec }^n}\phi {{\sec }^2}\phi }}}= 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^n}\phi d\phi } = {I_n}$ Now by integrating by parts with ${\cos ^{n - 1}}\phi= u$ $\cos \phi d\phi= dv$ we get that ${I_n}= \frac{{n - 1}}{n}{I_{n - 2}}$ Thus we get that ${I_n}= \frac{{n - 1}}{n}\frac{{n - 3}}{{n - 2}}\frac{{n - 5}}{{n - 4}} \cdots {I_0}$ for even $n$ ${I_n}= \frac{{n - 1}}{n}\frac{{n - 3}}{{n - 2}}\frac{{n - 5}}{{n - 4}} \cdots {I_1}$ for odd $n$ Now ${I_0}= \frac{\pi }{2}$ ${I_1}= 1$ So writing $n$ as $2m$ or $2m+1$ we get that. ${I_n}= \frac{{2m}}{{2m + 1}}\frac{{2m - 2}}{{2m - 1}}\frac{{2m - 4}}{{2m - 3}} \cdots 1 = \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}}$ ${I_n}= \frac{{2m - 1}}{{2m}}\frac{{2m - 3}}{{2m - 2}}\frac{{2m - 5}}{{2m - 4}} \cdots \frac{\pi }{2} = \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2}$ which you can always write in terms of the Gamma function and factorials knowing that ${2^n}\sqrt \pi \Gamma \left( {n + \frac{1}{2}} \right)= \left( {2n - 1} \right)!!$
 December 29th, 2011, 07:29 PM #13 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals Why is the 3rd integral non zero if the function being integrated is even? My doubt arises since for any even function: $\int\limits_{ - a}^a {f\left( x \right)dx}= 0$
 December 29th, 2011, 07:30 PM #14 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals I'm sorry, I meant for odd functions, but the function in question is even, so never mind.
 December 29th, 2011, 08:00 PM #15 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals I'd like to provide a proof for $\int\limits_{ - \infty }^\infty {\frac{x}{{{x^2} + {a^2}}}\sin xdx}= \pi {e^{ - a}}$ Theorem $\int\limits_{ - \infty }^\infty {\frac{x}{{{x^2} + {a^2}}}\sin xdx}= \pi {e^{ - a}}$ Proof $\int\limits_{ - \infty }^\infty {\frac{x}{{{x^2} + {a^2}}}\sin xdx}= \phi \left( a \right)$ Substitute $x= at$ to get $\int\limits_{ - \infty }^\infty {\frac{t}{{{t^2} + 1}}\sin \left( {at} \right)dt}= \phi \left( a \right)$ Differentiating under Leibniz's Rule: $\int\limits_{ - \infty }^\infty {\frac{{{t^2}}}{{{t^2} + 1}}\cos \left( {at} \right)dt}= \phi #39;\left( a \right) = \varphi \left( a \right)$ Given that $\frac{{{t^2}}}{{{t^2} + 1}}= 1 - \frac{1}{{{t^2} + 1}}$ and that $\int\limits_{ - \infty }^\infty {\cos \left( {at} \right)dt}= \frac{1}{a}\int\limits_{ - \infty }^\infty {\cos \left( x \right)dx = } \frac{1}{a}\sum\limits_{v = - \infty }^\infty {\int\limits_{v\pi }^{\left( {v + 1} \right)\pi } {\cos xdx} = } \frac{1}{a}\sum\limits_{v = - \infty }^\infty {\sin \left( {v + 1} \right)\pi + \sin \pi v} = 0$ we have got $\int\limits_{ - \infty }^\infty {\frac{{{t^2}}}{{{t^2} + 1}}\cos \left( {at} \right)dt}= - \int\limits_{ - \infty }^\infty {\frac{1}{{{t^2} + 1}}\cos \left( {at} \right)dt} = \varphi \left( a \right)$ but again, by Lebniz's Rule: $\int\limits_{ - \infty }^\infty {\frac{t}{{{t^2} + 1}}\sin \left( {at} \right)dt}= \varphi #39;\left( a \right)$ so $\phi \left( a \right)$ satisfies the differential equation $\phi \left( a \right) + \phi '\left( a \right) = 0$ thus $\phi \left( a \right)= {c_1}{e^{ - a}}$ and since $\phi \left( 0 \right)= \pi$ we have $\phi \left( a \right)= \pi {e^{ - a}}$ QED Although this proof is not really rigorous, I like it.
 December 29th, 2011, 08:03 PM #16 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals And 2 follows from 3 since 2 is the negative integral of 3.
 December 30th, 2011, 06:33 PM #17 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals Ok here goes another one: $\int\limits_0^\infty {\frac{{\log x}}{{{x^2} + {a^2}}}dx= }$ $x= at$ $dx= adt$ $\frac{1}{a}\int\limits_0^\infty {\frac{{\log at}}{{{t^2} + 1}}dx= } \frac{1}{a}\int\limits_0^\infty {\frac{{\log t}}{{{t^2} + 1}}dx} + \frac{1}{a}\int\limits_0^\infty {\frac{{\log a}}{{{t^2} + 1}}dx} = \frac{1}{a}\int\limits_0^\infty {\frac{{\log t}}{{{t^2} + 1}}dx} + \frac{\pi }{{2a}}\log a =$ Now $\int\limits_0^\infty {\frac{{\log t}}{{{t^2} + 1}}dx}=$ $t= {e^x}$ $dt= {e^x}dx$ $\int\limits_{ - \infty }^\infty {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx}=$ $f\left( { - x} \right)= \frac{{ - x{e^{ - x}}}}{{{e^{ - 2x}} + 1}} = \frac{1}{{{e^{2x}}}}\frac{{ - x{e^{ - x}}{e^{2x}}}}{{{e^{ - 2x}} + 1}} = - \frac{{x{e^x}}}{{{e^{2x}} + 1}} = - f\left( x \right)$ Thus $\int\limits_{ - \infty }^\infty {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx}= 0$ Then $\int\limits_0^\infty {\frac{{\log x}}{{{x^2} + {a^2}}}dx}= \frac{\pi }{{2a}}\log a$
 January 2nd, 2012, 01:00 PM #18 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 More directly,           ? if I = ?  ln(x)/(x² + a²) dx, where a > 0,          0 let x = a²/u, dx = -a²/u² du.       ?                                                           ?                                     ? I = ? -ln(a²/u)/((a²/u)² + a²) (-a²/u²) du = ?  ln(a²/u)/(a² + u²) du = ?  ln(a²)/(a² + u²) du - I.       0                                                           0                                     0                                                        ? Hence I = (1/2)ln(a²)[(1/a)atan(u/a)]  = (1/a)ln(a)?/2.                                                        0
 January 2nd, 2012, 05:46 PM #19 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: Nine Integrals Nice one. I'm trying to polish my proof for 3.

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