Open Problem (Calculus - 1)

Evaluate the following improper integrals and sum of series:


xsinαxx4+x2+1 dx \Large\color{#3D99F6}{\int_{-\infty}^\infty\frac{x\sin\alpha x}{x^4+x^2+1}\ dx}


cosβxx6+1 dx \Large\color{#D61F06}{\int_{-\infty}^\infty\frac{\cos\beta x}{x^6+1}\ dx}


n=(1)nn2+γ2 \Large\color{#456461}{\sum_{n=-\infty}^\infty\frac{(-1)^n}{n^2+\gamma^2}}

Note by Tunk-Fey Ariawan
5 years, 3 months ago

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The third one is equal to πγcsch(γπ)\frac{\pi}{\gamma} \text{csch}\left(\gamma\pi\right), where csch is the hyperbolic cosecant. The first step is to take (1)nn2+γ2=k=0(1)n+kγ2kn2k+2\frac{(-1)^n}{n^2+\gamma^2} = \sum_{k=0}^{\infty} \frac{(-1)^{n+k} \gamma^{2k}}{n^{2k+2}} We can then switch the order to limak=0n=aa(1)n+kγ2kn2k+2\lim_{a\to\infty}\,\sum_{k=0}^{\infty} \sum_{n=-a}^{a} \frac{(-1)^{n+k} \gamma^{2k}}{n^{2k+2}} to get k=0(1)k1(24k)ζ(2k+2)γ2k\sum_{k=0}^{\infty} (-1)^{k-1} \left(2-4^{-k}\right) \zeta (2 k+2)\gamma^{2k} This of course neglects the (1)002+γ2\frac{(-1)^0}{0^2+\gamma^2} term (which is undefined for n=0n = 0 in the denominator of the summand), so we must add 1γ2\frac{1}{\gamma^2} to our previous sum to get 1γ2+k=0(1)k1(24k)ζ(2k+2)γ2k\frac{1}{\gamma^2} + \sum_{k=0}^{\infty} (-1)^{k-1} \left(2-4^{-k}\right) \zeta (2 k+2)\gamma^{2k} We can see the relation between these coefficients and the coefficients of the series for the hyperbolic cosecant to get that our answer should be πγcsch(γπ)\boxed{\frac{\pi}{\gamma} \text{csch}\left(\gamma\pi\right)}.

Michael Lee - 5 years, 3 months ago

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Excellent!! How about this one. ¨\ddot\smile

Tunk-Fey Ariawan - 5 years, 3 months ago

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Lol this is what I keep doing too - directing people to the IMPOSSIBRU integral xD

John Muradeli - 5 years, 2 months ago

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Actually, I got the correct answer to that one by hand-estimating. I'm fairly certain that your answer is not exactly correct. To 50 digits, the integral is 0.59738180945180348461311323509087376430643859042556, whereas to 50 digits, the right side is 0.59738519747762744103977711867133271207190293243263.

Michael Lee - 5 years, 3 months ago

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minyak

uzumaki nagato tenshou uzumaki - 4 years, 10 months ago

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Quick sketch : the first two can be done using complex analysis just consider the integral over a half circle from RR to R-R counterclockwise and (R,R)(-R,R).

For the sum, consider the integral of zπcsc(πz)z2+x2z\mapsto \frac{\pi \csc(\pi z)}{z^2+x^2} over a rectangle with vertices (n+1/2){1+i,1i,1+i,1i}(n+1/2)\left\{1+i,1-i,-1+i,-1-i\right\}.

I can't write full solutions now, but I'll write when have time.

Haroun Meghaichi - 5 years, 3 months ago

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+1 for your comment. I know how to evaluate them using residue method but I am interested in evaluating them using elementary methods. Anyway, could you please post your solution to this problem: OMG !! Zeta and Pi in one integral!. Thank you.

Tunk-Fey Ariawan - 5 years, 3 months ago

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@Tunk-Fey Ariawan Sorry, I do not have much time and the evaluation of ψ4(1/3)\psi_4(1/3) is lengthy. To simplify Pranav's result you have only to calculate ψ4(1/3)\psi_4(1/3) (using some basic polygamma identities).

Haroun Meghaichi - 5 years, 3 months ago

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@Haroun Meghaichi I did solve your problem in similar fashion as @Pranav Arora did. I thought you had a different approach since you commented to him: "Have you noticed that the fraction has a partial fraction decomposition that might help?"

Tunk-Fey Ariawan - 5 years, 3 months ago

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@Tunk-Fey Ariawan @Tunk-Fey Ariawan I've added a solution to that problem. you may see that it is long, that's why I couldn't post it until I had time.

Haroun Meghaichi - 5 years, 3 months ago

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@Haroun Meghaichi It look like we did it by using the same approach. Nice solution!

Tunk-Fey Ariawan - 5 years, 3 months ago

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Problem No 2


Answer is πeβ6+π3eβ/2(sin(3β2)+cos(3β2))\displaystyle \boxed{\frac{\pi e^{-\beta}}{6} + \frac{\pi}{\sqrt3}e^{-\beta/2}\left(\sin(\frac{\sqrt3 \beta}{2})+\cos(\frac{\sqrt3 \beta}{2})\right)}

Aman Rajput - 3 years, 9 months ago

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please anybody solve the integral using the residue theorem

Hasan Kassim - 5 years, 2 months ago

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Problem No 1


Answer is

2π3e3α2sin(α2)\displaystyle \boxed{\frac{2\pi}{\sqrt3}e^{-\frac{\sqrt3 \alpha}{2}}\sin(\frac{\alpha}{2})}

Aman Rajput - 3 years, 9 months ago

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