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$Version "10.0 for Mac OS X x86 (64-bit) (September 10, 2014)"$Assumptions = {a > 1}; Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], {x, 0, ArcCosh[a]}] Pi/2 f[a_?NumericQ] := Module[ {ar = Rationalize[a, 0]}, NIntegrate[ 1/Sqrt[-1 + ar^2*Sech[x]^2], {x, 0, ArcCosh[ar]}, WorkingPrecision -> 20] // Chop] ...

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It seems to be a branch cut issue. It depends when the substitution is done, that causes the integrator to go one way vs. the other. This below shows difference ClearAll[x, a]; sol = Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], x] The integrand at x=0 is always zero, so we can ignore this, The issue is with the upper limit low = Limit[sol, x -> 0] (*0*) ...

1

Your syntax is wrong for Assuming Clear[fx]; fx[x_] := Piecewise[{ {(1/2) a*E^(-a*x), x >= 0}, {(1/2) a*E^(a*x), x < 0}}] Assuming[ {x \[Element] Reals && a \[Element] Reals && a > 0 && t \[Element] Reals}, Integrate[fx[x], {x, -Infinity, t}]] Piecewise[{{E^(a*t)/2, t <= 0}}, ((1/2)*(-1 + ...

4

To begin with, in the definition of fx include the parameter "a" as an argument and write x_ instead of x : fx[x_, a_] := Piecewise[{{(1/2) a*E^(-a*x), x >= 0}, {(1/2) a*E^(a*x), x < 0}}] If you now you tell Mathematica that "t" is real and that a>0 there is no problem integrating fx: Integrate[fx[x, a], {x, -Infinity, t}, Assumptions -> {a ...

2

I would use a slightly different derivation but the concept is basically the same. Here $I_q:=4\pi \int_0^{\infty}P(r)\frac{\sin qr}{qr}dr$. Hence $qI_q=4\pi \int_0^{\infty}P(r)\frac{\sin qr}{r}dr$. The derivative of $qI_q$ with respect to $q$ is $$\frac{d}{dq}(qI_q)=4\pi\int_{0}^{\infty}P(r)\cos(qr)\,dr.$$ This relation can be ...

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First: There might be a typo in your expression as b2 and c2 do not appear. This is the solution for the expression like it is written in your question: When you have more information on parameters, you can pass it to the Integrate function in the Assumptions option: Integrate[..., Assumptions -> b1 > 0 && b2 > 0 && c1^2 >= 0 ...

3

This answer contains no Mathematica, only mathematics. Let $P$ be extended to negative input via $P(-x)=P(x)$, and let $k_q(x) = \sin(qx)/qx = \operatorname{sinc}\bigl(\frac q\pi x\bigr)$. Both $P$ and $k_q$ are even functions. You have \begin{align} I_q &= 4\pi\int_0^\infty P(r)k_q(r)\,\mathrm dr \\ &= 2\pi\int_{-\infty}^\infty ...

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