MHB Evaluating an Integral with Exponential Factors

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The integral of interest is shown to equal \(\frac{e^{a^{4}b^{4}}}{4a} \Gamma \left( \frac{1}{4} \right)\) for \(a > 0\) and \(b \ge 0\). The approach involves integrating the function \(f(z) = e^{-a^{4}z^{4}}\) around a rectangular contour in the complex plane. By applying the residue theorem, the contributions from the contour lead to a simplification where the real parts are equated. The limit of certain integrals vanishes as \(R\) approaches infinity, confirming the equality. This method effectively demonstrates the evaluation of the integral with exponential factors.
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Show that for $a >0$ and $b \ge 0$,

$$\int_{0}^{\infty} e^{-a^{4}x^{2}(x^{2}-6b^{2})} \cos \Big(4a^{4}bx(x^{2}-b^{2}) \Big) \ dx = \frac{e^{a^{4}b^{4}}}{4a} \Gamma \left( \frac{1}{4} \right) $$
 
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Hint:

Integrate $ \displaystyle f(z)=e^{-a^{4}z^{4}}$ around the appropriate rectangular contour.
 
Hint 2:

Integrate $ \displaystyle e^{-a^{4}z^{4}}$ around a rectangle with vertices at $z=0, z= \infty, z= \infty + ib$, and $z=ib$, and notice that the integral along the left side of the rectangle is purely imaginary and that the integral along the right side of the rectangle vanishes.
 
Let $f(z) = e^{-a^{4}z^{4}}$ and integrate around a rectangle with vertices at $z= 0, z = \infty, z=\infty+ib$, and $z= ib$.Then going around the contour counterclockwise we have

$$\int_{0}^{\infty} e^{-a^{4}x^{4}} \ dx + \lim_{R \to \infty} \int_{0}^{b} f(R + it) \ i dt - \int_{0}^{\infty} f(t+ib) \ dt - \int_{0}^{b} f(it) i \ dt = 2 \pi i (0)=0 \ \ (1)$$$$\int_{0}^{\infty} e^{-a^{4}x^{4}} \ dx = \frac{1}{4a} \int_{0}^{\infty} e^{-u} u^{1/4-1} \ du = \frac{1}{4a} \Gamma \left(\frac{1}{4} \right) $$$$\Big| \int_{0}^{b} f(R+it) i \ dt \Big| \le \int_{0}^{b} \Big| e^{-a^{4}(R+it)^{4}} \Big| \ dt = e^{-a^{4}R^{4}} \int_{0}^{b} e^{6a^{4}R^{2}t^{2}} e^{-a^{4}t^{4}} \ dt$$

$$ < e^{-a^{4} R^{4}} \int_{0}^{b} e^{6a^{4} R^{2} b^{2}} e^{-a^{4}b^{4}} \ dt = b e^{-a^{4}R^{2}(R^{2}-6b^{2})} e^{-a^{4}b^{4}} \to 0 \ \text{as} \ R \to \infty$$$$ \int_{0}^{\infty} f(t+ib) \ dt = e^{-a^{4}b^{4}} \int_{0}^{\infty} \exp \Big(-a^{4}(t^{4}+4it^{3}b-6t^{2}b^{2}-4itb^{3}) \Big) \ dt$$$$ \int_{o}^{b} f(it) \ dt = i \int_{0}^{b}e^{-a^{4}t^{4}} \ dt $$Now equate the real parts on both sides of (1).$$ \frac{1}{4a} \Gamma \left(\frac{1}{4} \right) - e^{-a^{4}b^{4}} \int_{0}^{\infty} e^{-a^{4}t^{4} -6a^{4}t^{2}b^{2}} \cos \Big(4a^{4}t^{3}b - 4a^{4}tb^{3} \Big) \ dt = 0 $$

$$ \implies \int_{0}^{\infty} e^{-a^{4}t^{2}(t^{2}-6b^{2})} \cos \Big(4a^{4}bt(t^{2}-b^{2}) \Big) \ dx = \frac{e^{a^{4}b^{4}}}{4a} \Gamma \left( \frac{1}{4} \right) $$
 
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