Verified answer

I'll compute the general integral.

Partial fractions and using roots of unity gives us

1/(t^n + 1) = (-1/n) Σ(k=1 to n) a/(t - a), where a = e^((2k-1)πi/n).

Since this is given in complex numbers, we pair off complex conjugate pairs of terms to rewrite this as a real-valued function.

To do this neatly, write the sum in reverse (reindex k to n-k+1):

1/(t^n + 1) = (-1/n) Σ(k=1 to n) e^((2(n-k+1)-1)πi/n) / (t - e^((2(n-k+1)-1)πi/n))

.................= (-1/n) Σ(k=1 to n) e^((-2k+1)πi/n) / (t - e^((-2k+1)πi/n)), since e^(2πi) = 1

.................= (-1/n) Σ(k=1 to n) a*/(t - a*), where a* = e^(-(2k-1)πi/n)), the conjugate of a.

Adding this to the first version of 1/(t^n + 1) and dividing by 2 yields

1/(t^n + 1) = (-1/(2n)) Σ(k=1 to n) [a/(t - a) + a*/(t - a*)].

Next, note that

a/(t - a) + a*/(t - a*), where a* = conjugate of e^((2k-1)πi/n) = e^(-(2k-1)πi/n)

= [a(t - a*) + a*(t - a)] / [(t - a)(t - a*)]

= ((a+a*)t - 2aa*) / (t² - (a+a*)t + aa*)

= (2 cos((2k-1)π/n) t - 2 * 1) / (t² - 2 cos((2k-1)π/n) t + 1), via e^(ix) = cos x + i sin x.

= 2(bt - 1)/(t² - 2bt + 1), where b = cos((2k-1)π/n).

Thus, 1/(t^n + 1)

= (-1/n) Σ(k=1 to n) (bt - 1)/(t² - 2bt + 1)

= (-1/n) Σ(k=1 to n) (bt - 1)/((t - b)² + 1 - b²), completing the square

= (-1/n) Σ(k=1 to n) (b(t - b) - (1 - b²))/((t - b)² + 1 - b²)

Finally, ∫ dt/(t^n + 1)

= (-1/n) Σ(k=1 to n) [(b/2) ln((t - b)² + 1-b²) - ((1 - b²)/√(1 - b²)) arctan((t - b)/√(1 - b²))] + C

= (-1/(2n)) Σ(k=1 to n) [b ln(t² - 2bt + 1) - 2√(1 - b²) arctan((t - b)/√(1 - b²))] + C

Since the integral converges for n > 1, we have

∫(-∞ to ∞) dt/(t^n + 1)

= lim(s→∞) ∫(-s to s) dt/(t^n + 1)

= (-1/(2n)) Σ(k=1 to n) [0 - 2√(1 - b²) (π/2 - (-π/2))], via quotient rule for logs

= (π/n) Σ(k=1 to n) |sin((2k-1)π/n)|, since b = cos((2k-1)π/n).

Due to the symmetry of sine in the middle of one period, we can rewrite this as

∫(-∞ to ∞) dt/(t^n + 1) = (2π/n) Σ(k=1 to [n/2]) sin((2k-1)π/n),

where [ ] denotes the floor (greatest integer) function.

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Double checks:

n = 2 ==> (2π/2) Σ(k=1 to 1) sin((2k-1)π/2) = π * 1 = π,

and ∫(-∞ to ∞) dt/(t^2 + 1) = arctan t {from -∞ to ∞} = π.

n = 3 ==> (2π/3) Σ(k=1 to 1) sin((2k-1)π/3) = (2π/3) (√3/2) = π/√3,

and ∫(-∞ to ∞) dt/(t^3 + 1) = π/√3 (from the other post).

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Link:

https://math.stackexchange.com/questions/1354106/w...

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I hope this helps!

∫ dt/(t^3+1)

= ∫ dt/((t+1)(t^2+t+1))

1/((t+1)(t^2-t+1)) = A/(t+1) + (Bt+C) /(t^2-t+1)

multiply both sides by (t+1)(t^2-t+1)

1 = A(t^2-t+1) + (Bt+C)(t+1)

Let t=-1

1 = A((-1)^2-(-1)+1) + 0

1 = 3A

A = 1/3

Match the coefficients of t^2

0 = A+B

B = -A

B = -1/3

Match the coefficients of t

0 = -A+B+C

0 = -1/3-1/3+C

C = 2/3

1/((t+1)(t^2-t+1)) = (1/3) ∫ dt /(t+1) - (1/3) ∫ t dt /(t^2-t+1) +(2/3) ∫ dt /(t^2-t+1)

(1/3) ∫ dt /(t+1) = (1/3) ln |t+1| ----------(1)

Consider

∫ t dt / (t^2-t+1)

t = (1/2) (2t-1+1)

∫ t dt /(t^2-t+1) = (1/2)∫ (2t-1) dt/(t^2-t+1) + (1/2) ∫ 1 dt /(t^2-t+1)

(1/2)∫ (2t-1) dt/(t^2-t+1)

Let y = t^2-t+1

dy=(2t-1) dt

(1/2) ∫ (2t-1)dt/(t^2-t+1) = (1/2)∫ dy/y = (1/2) ln|y| = (1/2) ln (t^2-t+1)

∫ dt/(t^2-t+1) = ∫ dt/ ((t-1/2)^2+3/4)

Let u= t-1/2

du = dt

∫ dt/((t-1/2)^2 +3/4) = ∫ du/(u^2+3/4)

Factor out 3/4 from the denominator

= (4/3) ∫ du /(4u^2/3+ 1)

= (4/3) ∫ du /((2u/√3)^2+1)

Let w= 2u/√3

dw = 2du/√3

du = (√3/2) dw

(4/3) ∫ dw /((2u/√3)^2+1) = (3/4)(√3/2) ∫ dw /(w^2+1)

= (4/3)(√3/2) tan^-1(w)

= (2/√3) tan^-1(2u/√3)

= (2/√3) tan^-1(2(t-1/2)/√3)

= (2/√3) tan^-1( (2t-1)/√3)

Thus:

∫ dt/(t^2-t+1) = (2/√3) tan^-1((2t-1)/√3)

(1/2) ∫ 1 dt /(t^2-t+1) = (1/2)(2/√3) tan^-1((2t-1)/√3) =(1/√3) tan^-1((2t-1)/√3)

∫ t dt / (t^2-t+1) = (1/2) ln (t^2-t+1) + (1/√3) tan^-1(((2t-1)/√3)

(-1/3) ∫ t dt / (t^2-t+1) = (-1/6) ln (t^2-t+1) + (-1/(3√3)) tan^-1((2t-1)/√3) ----------(2)

(2/3) ∫ dt /(t^2-t+1) = (2/3)(2/√3) tan^-1( (2t-1)/√3) = (4/(3√3)) tan^-1((2t-1)/√3) ---------- (3)

∫ dt/(t^3+1) = (1) + (2) + (3)

= (1/3) ln |t+1|- (1/6) ln (t^2-t+1) + (-1/(3√3)) tan^-1(((2t-1)/√3) + (4/(3√3)) tan^-1((2t-1)/√3)

= (1/3) ln |t+1|- (1/6) ln (t^2-t+1) + (1/√3)) tan^-1((2t-1)/√3)

Let F(t) = (1/3) ln |t+1|- (1/6) ln (t^2-t+1) + (1/√3)) tan^-1((2t-1)/√3)

lim t-->∞ (1/3) ln |t+1|- (1/6) ln (t^2-t+1)

= (2/6) ln(t+1) - (1/6) ln(t^2-t+1)

= (2 ln(t+1) - ln(t^2-t+1))/6

= (1/6) lim t-->∞ (ln ((t+1)^2 /(t^2-t+1))

= (1/6) ln (lim t-->∞ ((t^2+2t+1)/(t^2-t+1)))

Since t^2+2t+1 and t^2-t+1 are both of degree 2, the ratio approaches 1 as t -->∞

= (1/6) ln (1) = 0

lim t-->∞ (1/√3)) tan^-1(((2t-1)/√3) = (1/√3) tan^-1(∞ )

= (1/√3)(pi/2)

= pi/(2√3)

Therefore, the integral has the limit 0 +pi/(2√3) as t approaches ∞

The integral has the limit -pi/(2√3) as t approaches -∞

Lim t-->∞ F(∞ )-F(-∞ ) = pi/√3