Evaluate the following improper integrals and sum of series:

\[ \Large\color{blue}{\int_{-\infty}^\infty\frac{x\sin\alpha x}{x^4+x^2+1}\ dx} \]

\[ \Large\color{red}{\int_{-\infty}^\infty\frac{\cos\beta x}{x^6+1}\ dx} \]

\[ \Large\color{darkgreen}{\sum_{n=-\infty}^\infty\frac{(-1)^n}{n^2+\gamma^2}} \]

## Comments

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TopNewestThe third one is equal to \(\frac{\pi}{\gamma} \text{csch}\left(\gamma\pi\right)\), where csch is the hyperbolic cosecant. The first step is to take \[\frac{(-1)^n}{n^2+\gamma^2} = \sum_{k=0}^{\infty} \frac{(-1)^{n+k} \gamma^{2k}}{n^{2k+2}}\] We can then switch the order to \[\lim_{a\to\infty}\,\sum_{k=0}^{\infty} \sum_{n=-a}^{a} \frac{(-1)^{n+k} \gamma^{2k}}{n^{2k+2}}\] to get \[\sum_{k=0}^{\infty} (-1)^{k-1} \left(2-4^{-k}\right) \zeta (2 k+2)\gamma^{2k}\] This of course neglects the \(\frac{(-1)^0}{0^2+\gamma^2}\) term (which is undefined for \(n = 0\) in the denominator of the summand), so we must add \(\frac{1}{\gamma^2}\) to our previous sum to get \[\frac{1}{\gamma^2} + \sum_{k=0}^{\infty} (-1)^{k-1} \left(2-4^{-k}\right) \zeta (2 k+2)\gamma^{2k}\] 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 \(\boxed{\frac{\pi}{\gamma} \text{csch}\left(\gamma\pi\right)}\). – Michael Lee · 2 years, 5 months ago

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one. \(\ddot\smile\) – Tunk-Fey Ariawan · 2 years, 5 months ago

Excellent!! How about thisLog in to reply

– John Muradeli · 2 years, 4 months ago

Lol this is what I keep doing too - directing people to the IMPOSSIBRU integral xDLog in to reply

– Michael Lee · 2 years, 5 months ago

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.Log in to reply

– Uzumaki Nagato Tenshou Uzumaki · 2 years ago

minyakLog in to reply

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

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

I can't write full solutions now, but I'll write when have time. – Haroun Meghaichi · 2 years, 5 months ago

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OMG !! Zeta and Pi in one integral!. Thank you. – Tunk-Fey Ariawan · 2 years, 5 months ago

+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:Log in to reply

@Tunk-Fey Ariawan Sorry, I do not have much time and the evaluation of \(\psi_4(1/3)\) is lengthy. To simplify Pranav's result you have only to calculate \(\psi_4(1/3)\) (using some basic polygamma identities). – Haroun Meghaichi · 2 years, 5 months ago

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@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 · 2 years, 5 months ago

I did solve your problem in similar fashion asLog in to reply

@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 · 2 years, 5 months ago

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– Tunk-Fey Ariawan · 2 years, 5 months ago

It look like we did it by using the same approach. Nice solution!Log in to reply

Problem No 2Answer is \[\displaystyle \boxed{\frac{\pi e^{-\beta}}{6} + \frac{\pi}{\sqrt3}e^{-\beta/2}\left(\sin(\frac{\sqrt3 \beta}{2})+\cos(\frac{\sqrt3 \beta}{2})\right)}\] – Aman Rajput · 11 months ago

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Problem No 1Answer is

\[\displaystyle \boxed{\frac{2\pi}{\sqrt3}e^{-\frac{\sqrt3 \alpha}{2}}\sin(\frac{\alpha}{2})}\] – Aman Rajput · 11 months ago

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please anybody solve the integral using the residue theorem – Hasan Kassim · 2 years, 4 months ago

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