Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need a little help with this question, please!

I have to evaluate the real convergent improper integrals using RESIDUE THEORY (vital that I use this), using the following contour:

$$\int\limits_0^\infty \frac{\log x} {(1+x^2)^2} dx$$

Using this contour:

enter image description here

$R>1$ and $r<1$

share|improve this question
    
I edited. Is the expression correct? –  Pedro Tamaroff Mar 7 '12 at 18:06
    
Why is it "vital" to use one particular tool? If it's a homework question, then please tag it appropriately. –  Henning Makholm Mar 7 '12 at 18:49
    
Bany, try improving your accept rate! You can accept anwers by cilcking on the tick on the top left of them. –  Pedro Tamaroff Apr 6 '12 at 22:52
add comment

1 Answer 1

I'll give my humble idea to show the integral is $-\dfrac{\pi}{4}$.

With a change of variables ($x=e^u$) we have that

$$\mathcal{I}=\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } \int\limits_{ - \infty }^\infty {\frac{{u{e^u}}}{{{{\left( {1 + {e^{2u}}} \right)}^2}}}du} $$

We can write this as

$${\mathcal I} = \int\limits_{ - \infty }^\infty {\frac{{u{e^{ - u}}}}{{{{\left( {{e^{ - u}} + {e^u}} \right)}^2}}}du} $$

Putting $u=-v$ we have that

$${\mathcal I} = \int\limits_{ - \infty }^\infty {\frac{{u{e^{ - u}}}}{{{{\left( {{e^{ - u}} + {e^u}} \right)}^2}}}du} = -\int\limits_{ - \infty }^\infty {\frac{{v{e^v}}}{{{{\left( {{e^{ - v}} + {e^v}} \right)}^2}}}dv} $$

This means that

$$2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } \int\limits_{ - \infty }^\infty {\frac{{u\left( {{e^{ - u}} - {e^u}} \right)}}{{{{\left( {{e^u} + {e^{ - u}}} \right)}^2}}}du} $$

We can write this in terms of the hiperbolic functions, to get

$$2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{1}{2}\int\limits_{ - \infty }^\infty {\frac{{u\sinh u}}{{\cosh^2 u}}du} $$

Integration by parts gives ($(\operatorname{sech} u)'=-\dfrac{{\sinh u}}{{\cosh^2 u}}$)

$$ - \int\limits_{ - \infty }^\infty {\frac{{\sinh udu}}{{{{\cosh }^2}u}}} = \left[ {u\operatorname{sech} u} \right]_{ - \infty }^\infty - \int\limits_{ - \infty }^\infty {\frac{{du}}{{\cosh u}}} $$

Finally, you can easily check that

$$\int\limits_{ - \infty }^\infty {\frac{{du}}{{\cosh u}}} = \pi $$

and that $u \operatorname{sech} u$ is odd so the first term in the RHS is zero. Thus

$$\eqalign{ & 2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{2} \cr & I = \int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{4} \cr} $$

share|improve this answer
    
Yep it is a homework question, sorry for the confusion –  Bany Mar 7 '12 at 18:54
    
@Bany Have you come up with any ideas on how to do it? –  Pedro Tamaroff Mar 7 '12 at 19:02
    
yea well the first step is to caclulate the reside, ive worked out that this is 1/i since the poles are (+/- i) of order 2; but only +i lies in the contour. now i need to work out limits of integrals but im lost at this part –  Bany Mar 7 '12 at 19:08
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.