Evaluating an Integral with Exponential Factors

[polygamma]

Show that for $a >0$ and $b \ge 0$,

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

 

[polygamma]

Hint:

Spoiler

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

 

[polygamma]

Hint 2:

Spoiler

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.

 

[polygamma]

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).$$ \frac{1}{4a} \Gamma \left(\frac{1}{4} \right) - e^{-a^{4}b^{4}} \int_{0}^{\infty} e^{-a^{4}t^{4} -6a^{4}t^{2}b^{2}} \cos \Big(4a^{4}t^{3}b - 4a^{4}tb^{3} \Big) \ dt = 0 $$

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

 

[CellCoree]

I am excited to see this integral being evaluated with exponential factors. This is an important step in understanding the behavior of exponential functions and their impact on integrals. The given expression is a definite integral, which means that it represents the area under the curve of the function being integrated. In this case, the function has both an exponential and a trigonometric factor, which makes it a challenging but interesting integral to evaluate.

To begin, let us rewrite the integral in a more convenient form by substituting $u = a^{4}x^{2}(x^{2}-6b^{2})$ and $dv = \cos \Big(4a^{4}bx(x^{2}-b^{2}) \Big)dx$. This gives us $du = 2a^{4}x(x^{2}-3b^{2})dx$ and $v = \frac{1}{4a^{4}b} \sin \Big(4a^{4}bx(x^{2}-b^{2}) \Big)$. Using the integration by parts formula, we can rewrite the integral as:

$$\int_{0}^{\infty} e^{-u} dv = \Big[e^{-u}v \Big]_{0}^{\infty} - \int_{0}^{\infty} vdu $$

Since $a > 0$, the limits of integration can be changed to $0$ and $+\infty$ without changing the value of the integral. Also, since $b \ge 0$, we know that $x^{2}-6b^{2}$ and $x^{2}-b^{2}$ are always positive, which means that $u$ and $v$ are well-defined. Using these simplifications, we can rewrite the integral as:

$$\int_{0}^{\infty} e^{-u} dv = \Big[e^{-u}v \Big]_{0}^{\infty} - \int_{0}^{\infty} vdu = \Big[e^{-u}v \Big]_{0}^{\infty} - \frac{1}{4a^{4}b} \int_{0}^{\infty} \sin \Big(4a^{4}bx(x^{2}-b^{2}) \Big) \cdot 2a^{4}x(x^{2