Help me to find a Definite Integral

sunveer
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1. Evaluate [PLAIN]http://www.goiit.com/equations/2011/5/25/795493e8-6a4d-42b0-8c18-2afa4e75b653.png[/b]



I have tried hard to solve it but I am not getting the exact answer.
The ans is given to be -a^2/(1+a^2)
 
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Integration by parts. Let u= e^{-2x}[/tex] and dv= cos(2ax)<br /> <br /> The result will involve <br /> \int e^{-2x}sin(2ax)dx&amp;lt;br /&amp;gt; so use integration by parts again. Let u= e^{-2x} again but let dv= sin(2ax).&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; That will give you a formula that again involves&amp;amp;lt;br /&amp;amp;gt; \int e^{-2x}cos(2ax)dx &amp;amp;lt;br /&amp;amp;gt; so set the whole thing equal to that integral.&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Try that and if you have a problem show your work here.
 
I solved it by Integration by parts
but
at last I am getting

(1+a^2){∫e−2xcos(2ax)dx}=0
 
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sunveer said:
1. Evaluate [PLAIN]http://www.goiit.com/equations/2011/5/25/795493e8-6a4d-42b0-8c18-2afa4e75b653.png[/b]



I have tried hard to solve it but I am not getting the exact answer.
The ans is given to be -a^2/(1+a^2)

Your integral is divergent. However, the integral with exp(-2*|x|) instead of exp(-2*x) does give 1/(1+a^2), and the integral with exp(-x^2) instead of exp(-2*x) gives sqrt(pi) * exp(-a^2).

RGV
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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