Consider this series $I_n$:
$$I_n = \int\limits_0^{\pi/4} \tan^{2n}(x)\ dx$$
How do I use Mathematica to simplify $I_n + I_{n+1}$ as a function of $n$?
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$\begingroup$ Mathematica input should be provided, images do not cut and paste very well. $\endgroup$– Daniel LichtblauCommented Mar 22, 2018 at 17:18
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1 Answer
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Usually it is convenient to use Assumptions, e.g.
Int[n_] = Integrate[Tan[t]^(2 n), {t, 0, Pi/4}, Assumptions -> n >= 0]
1/4 (-PolyGamma[0, 1/4 + n/2] + PolyGamma[0, 3/4 + n/2])
The result is immediate, because
FullSimplify[1/4 (-PolyGamma[0, 1/4 + n/2] + PolyGamma[0, 5/4 + n/2])]
1/(1 + 2 n)
i.e.
FullSimplify[ Int[n] + Int[n + 1]] // TraditionalForm
$$\frac{1}{2 n+1}$$
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1$\begingroup$ It is sufficient to assume
n>=0, i.e.,nneed not necessarily be an integer. $\endgroup$ Commented Mar 22, 2018 at 17:00 -
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$\begingroup$ Thanks ! Now, I need to show that In - I(n+1) <= 0. I use FullSimplify[Int[n] - Int[n + 1] >= 0] // TraditionalForm and I was expected a "True". But, nothing happens. Do you know why? $\endgroup$– nolwwwCommented Mar 22, 2018 at 17:58
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$\begingroup$ @nolw38 If you have another question, a good custom is asking it as a separate one. I would demonstrate in details how to prove it, however since you've neither upvoted nor accepted my answer I'll only suggest to look at the formula $5.9.16$ in NIST and use a simple inequality between arithmetic and geometric means. $\endgroup$– ArtesCommented Mar 22, 2018 at 19:19
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$\begingroup$ Sorry, just accepted your answer !! $\endgroup$– nolwwwCommented Mar 23, 2018 at 5:20