Lonely

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15 years, 13 days

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These are replies submitted by Lonely

thanks.

here $n$ is a positive number n >= 1.

i am interested in finding a closed form for the integral. is it possible:

thanks.

here $n$ is a positive number n >= 1.

i am interested in finding a closed form for the integral. is it possible:

Thanks. I want a formula that works for all n

But when i do

> `> `assuming`([diff(1/(1+x^2), `$`(x, n))], [n::posint]);

maple gives
                                /  1            \
                            diff|------, [x $ n]|
                                |     2         |
                                \1 + x          /
 

 

maple does not give

 

(n!/2)*(-exp(-(1/2)*n*ln(1+x^2))*sin(n*arctan(1, -x))*x+exp(-(1/2)*n*ln(1+x^2))*cos(n*arctan(1, -x))+exp(-(1/2)*n*ln(1+x^2))*sin(n*arctan(-1, -x))*x+exp(-(1/2)*n*ln(1+x^2))*cos(n*arctan(-1, -x)))/(1+x^2)

 

Thanks. I want a formula that works for all n

But when i do

> `> `assuming`([diff(1/(1+x^2), `$`(x, n))], [n::posint]);

maple gives
                                /  1            \
                            diff|------, [x $ n]|
                                |     2         |
                                \1 + x          /
 

 

maple does not give

 

(n!/2)*(-exp(-(1/2)*n*ln(1+x^2))*sin(n*arctan(1, -x))*x+exp(-(1/2)*n*ln(1+x^2))*cos(n*arctan(1, -x))+exp(-(1/2)*n*ln(1+x^2))*sin(n*arctan(-1, -x))*x+exp(-(1/2)*n*ln(1+x^2))*cos(n*arctan(-1, -x)))/(1+x^2)

 

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