Recent content by nshiell

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    Hardest Integral you will ever do

    yeah the r integral is not bad, but the theta integral is. Thats the part I can not get past.
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    Hardest Integral you will ever do

    yeah the keystone shape is a section of an annulus. Though the charge is not deposited at the origin of the annulus. its deposited at an arbitrary point.
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    Is the Convolution of Two Gaussians Always a Gaussian?

    int(exp(-x^2)*exp(-(r-x)^2))dx there is the limits are +/- infinity
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    Hardest Integral you will ever do

    1/rc is the RC time constant of the surface material. And yeah your rite the charge flows over the pads, and the pads detect it capacitively.
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    Hardest Integral you will ever do

    Okay here is the original problem. A cluster of charge is deposited onto a conductive surface and disperse like, p(r,t) = N*q/[(2*pi)*(2ht+w^2)]*exp[(-r^2)/(2*(2ht+w^2)] (This is the charge density) where h is 1/rc of the surface, w is the width of the cluster (unimportant here), N # of...
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    Is the Convolution of Two Gaussians Always a Gaussian?

    I've read on a bunch of websites that the convolution of two gaussians produces another gaussian however I have not seen this integration worked out. I am working on an integral which has a similar form as this convolution so it would be a great help too see. Does anyone know a book or website...
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    Hardest Integral you will ever do

    Okay so first the x is just another constant, I suppose it was a poor choice. And the order doesn't matter, which ever makes it easiest. And second, as for attempts i have made. The terms in the exponential are the result of a vector addition because the formal has to do with the distance from...
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    Hardest Integral you will ever do

    Hardest Integral you will ever do! int(exp(-a*r^2+a*r*x*cos(theta))*r)) This is a double integral over a keystone shape in polar coordinates. (so limits are finite, and not a complete circle). I've tried all sorts of things. I am not even sure if there is an analytic solution. Is there...