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Integral #2

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Show that for $a >0$ and $b \ge 0$,

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

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Hint:

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

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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).


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

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