Evaluate the following trIgonometric improper integral using the residue theorem

∫(-∞ to ∞) [sin(πx)cos(πx)/(2x^2 -x)]dx=-π
Can anyone help me with this problem please?

Note that ∫(-∞ to ∞) sin(πx)cos(πx) dx/(2x^2 - x)
= ∫(-∞ to ∞) (1/2) sin(2πx) dx/(2x^2 - x)
= Im [∫(-∞ to ∞) e^(2πix) dx/(4x^2 - 2x)].

*Note that the integrand has singularities at x = 0 and 1/2, which are on the x-axis.
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Fix R > 0, and 0 < ε < R/4 (or so), and consider ∫c e^(2πiz) dz/(4z^2 - 2z), where C is the closed contour (counterclockwise) consisting of Γ: upper half of |z| = R, the x-axis with x in [-R, ε] U [ε, 1/2 - ε] U [1/2 + ε, R], and γ, γ': upper halves of |z| = ε and |z - 1/2| = ε (with clockwise orientations).

So, ∫c e^(2πiz) dz/(4z^2 - 2z) = ∫Γ e^(2πiz) dz/(4z^2 - 2z) + ∫γ e^(2πiz) dz/(4z^2 - 2z)
+ ∫γ' e^(2πiz) dz/(4z^2 - 2z) + ∫(-R to -ε) e^(2πix) dx/(4x^2 - 2x)
+ ∫(ε to 1/2 - ε) e^(2πix) dx/(4x^2 - 2x) + ∫(1/2 - ε to R) e^(2πix) dx/(4x^2 - 2x)
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(i) ∫c e^(2πiz) dz/(4z^2 - 2z) = 0 by Cauchy's Theorem, because C encloses no singularities of the integrand (we indented the contour around z = 0 and z = 1/2).

(ii) lim(R→∞) ∫Γ e^(2πiz) dz/(4z^2 - 2z) = 0 by Jordan's Lemma.

(iii) Parameterize γ via z = εe^(it) with t in [0, π] and opposite orientation.

So, lim(ε→0+) ∫γ e^(2πiz) dz/(4z^2 - 2z)
= lim(ε→0+) ∫γ e^(2πiz) dz/((2z)(2z - 1))
= lim(ε→0+) -∫(t = 0 to π) e^(2πiεe^(it)) * iεe^(it) dt / (2εe^(it) (2εe^(it) - 1))
= lim(ε→0+) -∫(t = 0 to π) e^(2πiεe^(it)) * i dt / (2(2εe^(it) - 1))
= -∫(t = 0 to π) e^0 * i dt / (2(0 - 1))
= πi/2.
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(iv) Parameterize γ' via z = 1/2 + εe^(it) with t in [0, π] and opposite orientation.

So, lim(ε→0+) ∫γ' e^(2πiz) dz/(4z^2 - 2z)
= lim(ε→0+) ∫γ' e^(2πiz) dz/((2z)(2z - 1))
= lim(ε→0+) -∫(t = 0 to π) e^(2πi (1/2 + εe^(it))) * iεe^(it) dt / [2(1/2 + εe^(it)) (2(1/2 + εe^(it)) - 1)]
= lim(ε→0+) -∫(t = 0 to π) e^(2πi (1/2 + εe^(it))) * i dt / [2(1 + 2εe^(it))]
= -∫(t = 0 to π) e^(2πi (1/2 + 0)) * i dt / [2(1 + 0)]
= πi/2.
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So, letting R→∞ and ε→0+ (via (i) - (iv)) yields
0 = 0 + πi/2 + πi/2 + ∫(-∞ to ∞) e^(2πix) dx/(4x^2 - 2x)
==> ∫(-∞ to ∞) e^(2πix) dx/(4x^2 - 2x) = -πi

Take imaginary parts of both sides:
∫(-∞ to ∞) sin(πx)cos(πx) dx/(2x^2 - x) = -π, as required.

I hope this helps!