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 Yearling
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Feb
3
revised Compute the integrals using the residue theorem
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Feb
3
answered Compute the integrals using the residue theorem
Jan
31
comment Finding the value of this integral $ \int_{-\pi/4}^{\pi/4}{ (\cos{t} + \sqrt{1 + t^2} }\sin^3{t}\cos^3 {t})dt$?
Very nifty !...
Jan
31
comment Finding the value of this integral $ \int_{-\pi/4}^{\pi/4}{ (\cos{t} + \sqrt{1 + t^2} }\sin^3{t}\cos^3 {t})dt$?
You could split the integral into two terms. The first term should be easy to integrate. For the second, you may find $\frac{1}{2}\sin(2t)=\sin(t)\cos(t)$ useful.
Jan
30
comment why $\int{\cos({\pi}t)} dt = \frac{1}{\pi}\sin({\pi}t)$?
When something new is "added" we can assume the old rules will work, but often they don't work meaning there's more going on than we know. One way to think is to always remember that the rules we know at present are some special case of more general rules. So e.g. your statement $\int\cos(x)dx=\sin(x)+C$ is a special way of writing $\int\cos(1\cdot x)=\sin(x)+C$, but what if the $1$ were some other number $a$. What general rule is there to handle such cases? It's not as straight forward as the case when $a=1$.
Jan
27
comment Prove convergence of $\sum _{n=1}^{\infty }\sin(1/n)/n$
Since $1/n\to 0$ as $n\to\infty$, maybe you could consider $\sin(1/n)$ in a neighbourhood of $0$, e.g. $\sin(1/n)\to1/n$ as $n\to\infty$...
Jan
22
awarded  Yearling
Jan
22
comment Finding the result of an infinite sum
That's no reason to methodically downvote every element of a question though.
Jan
20
revised I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$
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Jan
19
answered I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$
Jan
19
revised I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$
added 2 characters in body
Jan
19
revised I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$
added 2 characters in body
Jan
19
comment Name for “3D quadrilateral” shape?
Thank you Andrew. I will have a think about your comment.
Jan
19
comment Name for “3D quadrilateral” shape?
Just found this: I think maybe polyhedron might cover it (in general). A polyhedron with 6 faces is called a hexahedron. mathworld.wolfram.com/Hexahedron.html
Jan
19
revised Name for “3D quadrilateral” shape?
added 62 characters in body
Jan
19
asked Name for “3D quadrilateral” shape?
Jan
19
revised How to evaluate the integral $\int_{0}^{\infty}\frac{\cos {(ax)}-\cos{(b x)}}{x^2 }dx$?
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Jan
19
answered How to evaluate the integral $\int_{0}^{\infty}\frac{\cos {(ax)}-\cos{(b x)}}{x^2 }dx$?
Jan
19
comment How to evaluate the integral $\int_{0}^{\infty}\frac{\cos {(ax)}-\cos{(b x)}}{x^2 }dx$?
Can you not use a keyhole contour?
Jan
19
comment How to evaluate the integral $\int_{0}^{\infty}\frac{\cos {(ax)}-\cos{(b x)}}{x^2 }dx$?
Another form is $$-\frac{1}{2}I=\int_0^\infty \frac{\sin\left(\frac{1}{2}(a+b)x\right)\sin\left(\frac{1}{2}(a-b)x\right)}{x^2}‌​{dx}.$$