# Tag Info

## New answers tagged series-expansion

0

lst = Table[{a, NIntegrate[(-(a^2/2) + x (-x + Sqrt[a^2 + x^2]))* Tanh[\[Pi]*x], {x, 0, \[Infinity]}]}, {a, 0.2, 1, 0.05}]; ListPlot[lst]

3

This is at least how I might start such a problem: First define a function that calculates the numerical integral (using your definition) from 0 to some number: res[a_?NumericQ, xmax_?NumericQ] := res[a, xmax] = NIntegrate[ x Tanh[Pi x] Sqrt[x^2 + a^2] - (a^2/2 + x^2) Tanh[\[Pi] x], {x, 0, xmax}, WorkingPrecision -> 50] You might notice ...

1

Is this is what you are after? Series[Sum[(-1)^n/QPochhammer[q^3, q^3, n], {n, 0, 10}], {q, 0, 10}] // Normal (* 1 + q^6 + q^9 *)

18

The standard built-in logarithm function is defined for complex variables as follows: Log[z] = Log[Abs[z]] + I Arg[z] The location of the branch cut is simply caused by the convention that polar angles of z are assumed to be in the range $-\pi$ to $\pi$. This same branch cut is also part of the definition of the built-in Arg function. Here is a different ...

8

The problem is that the real part of the argument in the Gamma functions is not recognized as being real. The only thing you have to do in order to circumvent this problem is to give a name to the quantity that appears as the real part and specify explicitly that it is real. Here I do this: Simplify[ Assuming[ y ∈ Reals && m ∈ Reals && q ...

3

The only thing wrong with your code is that Series doesn't return an expression suitable for evaluation at specific values of the expansion variable. You first have to convert the SeriesData object to a normal expression using Normal. This is the only change needed: i1[ϵ_] = Normal[Series[i[ϵ], {ϵ, 0, 1}]]; With this, the integration works as expected: ...

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