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8

One can simply use: Limit[((1 + a x)^(1/4) - (1 + b x)^(1/4))/x, x -> 0] to get the result. However Limit does know about the l'Hopital rule. Nonetheless there are different ways to go, e.g. let's use Series to write down the first few terms of the Taylor series of ((1 + a x)^(1/4) - (1 + b x)^(1/4))/x: Series[((1 + a x)^(1/4) - (1 + b x)^(1/4))/x, ...


5

This ordering is used in all functions in Mathematica, not just Integrate. Here's an example with Table: In[1]:= Table[f[i, j], {i, 3}, {j, 4}] Out[1]= {{f[1, 1], f[1, 2], f[1, 3], f[1, 4]}, {f[2, 1], f[2, 2], f[2, 3], f[2, 4]}, {f[3, 1], f[3, 2], f[3, 3], f[3, 4]}} The inner loop is according to j, the outer according i. In[2]:= ...


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 ...



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