Gamma , Γ, everywhere!

$\begin{aligned} I & = \int_0^\infty e^{-x^7} \blue{\sin x^7}\ dx & \small \blue{\text{By Euler's formula: }e^{\theta i} = \cos \theta + i\sin \theta} \\ & = \blue \Im \left(\int_0^\infty e^{-x^7 + \blue {x^7i}} \ dx \right) & \small \blue{\Im(\cdot) \text{ denotes the imaginary part function.}} \\ & = \Im \left(\int_0^\infty \frac {t^{-\frac 67}e^{-t}}{7\sqrt[7]{1-i}} dt \right) & \small \blue{\text{Let }t = (1-i)x^7 \implies dt = 7(1-i)x^6\ dx} \\ & = \Im \left(\frac \blue{\Gamma \left(\frac 17 \right)}{\blue 7 \red{\sqrt[7]{(1-i)}}} \right) & \small \blue{\text{Gamma function }\Gamma (s) = \int_0^\infty t^{s-1}e^{-t} dt} \\ & = \Im \left(\frac \blue{\Gamma \left(\frac 87 \right)}\red{\sqrt[7]{\sqrt 2 e^{\left(2n-\frac 14\right)\pi i}}} \right) & \small \blue{\text{and }\Gamma (1+s) = s \Gamma(s)} \\ & = \Im \left(\frac {e^{\frac {8n+1}{28} \pi i}\Gamma \left(\frac 87\right)}{\sqrt[14]2} \right) & \small \red{\text{By Euler's formula}} \\ & = \Im \left(\frac {\left(\cos \frac {(8n+1)\pi}{28} + i\sin \frac {(8n+1)\pi}{28} \right) \Gamma \left(\frac 87\right)}{\sqrt[14]2} \right) \\ & = \frac {\blue{\sin \frac {(8n+1)\pi}{28}}\Gamma \left(\frac 87\right)}{\sqrt[14]2} & \small \blue{\text{Checking with the numerical value of }I} \\ & = \frac {\blue{\sin \frac \pi {28}} \Gamma \left(\frac 87\right)}{\sqrt[14]2} \end{aligned}$

Therefore $\dfrac {a(b-c)}q = \dfrac {28(8-7)}{14} = \boxed 2$ .

$\int_0^∞ e^{-x^7} \sin(x^7)dx = \frac{1}{2i}(\int_0^∞ e^{-x^7 +ix^7} - \int_0^∞ e^{-x^7-ix^7}dx)$ As $sinx= \frac{e^{ix}-e^{-ix}}{2i}$

Take $x^7 = u$ then the integral becomes $\frac{1}{14i}(\int_0^∞ u^{-6/7}e^{-u(1-i)} du-\int_0^∞ u^{-6/7} e^{-u(1+i)} du)$ Take $u(1-i) = t , and , u(1+i) = p$ Then the integral becomes $\frac{1}{14i(1-i)^{1/7}} \int_0^∞ t^{-6/7} e^{-t}dt -\frac{1}{14i(1+i)^{1/7}} \int_0^∞ p^{-6/7} e^{-p}dp$ Which is $Γ(\frac{1}{7} ) (\frac{1}{14i(1-i)^{1/7}} - \frac{1}{14i(1+i)^{1/7}} )$ $(1-i)^{1/7} = \sqrt[14]{2} e^{-πi/28} , (1+i)^{1/7} = \sqrt[14]{2} e^{πi/28}$ So the integral will be $\frac{1}{14i(\sqrt[14]{2})}Γ(1/7)(2i \sin(\frac{π}{28}))$

Answer is $\boxed{\frac{Γ(\frac{8}{7}) \sin(\frac{π}{28})}{\sqrt[14]{2}}}$

$I = \int_{0}^{\infty} \mathrm{e}^{-x^7} \sin{x^7} \ dx$

Let: $x^7=z$ . This transforms the integral to:

$I = \frac{1}{7}\int_{0}^{\infty} \mathrm{e}^{-z} \sin{z} \ z^{-6/7} \ dz$

Knowing that:

$\sin{z} = \frac{\mathrm{e}^{iz} - \mathrm{e}^{-iz}}{2i}$

$\implies I =\frac{1}{7}\int_{0}^{\infty} \mathrm{e}^{-z} \left( \frac{\mathrm{e}^{iz} - \mathrm{e}^{-iz}}{2i}\right) \ z^{-6/7} \ dz$ $\implies I = \frac{1}{14i}\left(\int_{0}^{\infty} \mathrm{e}^{-(1-i)z} \ z^{-6/7} \ dz - \int_{0}^{\infty} \mathrm{e}^{-(1+i)z} \ z^{-6/7} \ dz\right)$

Let:

$J (a)= \int_{0}^{\infty} \mathrm{e}^{-az} \ z^{-6/7} \ dz$ $\implies I =\frac{1}{14i}\left(J(1-i) -J(1+i)\right)$

Taking $az = t$ transforms the integral to:

$J(a) = \frac{1}{a^{1/7}}\int_{0}^{\infty} \mathrm{e}^{-t} \ t^{-6/7} \ dt$ $J(a) = \frac{\Gamma\left(\frac{1}{7}\right)}{a^{1/7}}$

The above is obtained using the definition of the Gamma function. Plugging the above result in $I$ gives:

$I = \frac{1}{14i}\left(J(1-i) -J(1+i)\right)$ $I = \frac{\Gamma\left(\frac{1}{7}\right)}{7(2i)}\left( \frac{1}{(1-i)^{1/7} }- \frac{1}{(1+i)^{1/7}}\right)$ $I = \frac{\Gamma\left(\frac{8}{7}\right)}{2i}\left( \frac{1}{(1-i)^{1/7} }- \frac{1}{(1+i)^{1/7}}\right)$ $\because \Gamma(n+1) = n \Gamma(n)$

Now: $1-i = \sqrt{2} \mathrm{e}^{-i\pi/4}$ $(1-i)^{1/7} = 2^{1/14} \mathrm{e}^{-i\pi/28}$

Similarly:

$(1+i)^{1/7} = 2^{1/14} \mathrm{e}^{i\pi/28}$

Plugging these results and simplifying:

$I = \frac{\Gamma\left(\frac{8}{7}\right)}{2i}\left(\mathrm{e}^{i\pi/28} - \mathrm{e}^{-i\pi/28}\right)\frac{1}{2^{1/14}}$ $\implies I = \frac{\Gamma\left(\frac{8}{7}\right)\sin\left(\frac{\pi}{28}\right)}{2^{1/14}}$

Consider the integral, with $n > 0$ ,

$\begin{aligned} I_n = \int_{0}^{\infty} e^{-x^{n}} \sin{x^{n}} \, dx . \end{aligned}$

Substitute with $t = x^n$ , we have

$\begin{aligned} I_n = \frac{1}{n} \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{-t} \sin{t} \, dt \end{aligned}$

Define

$\begin{aligned} J_n^1(z) = \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{-t} \sin{zt} \, dt \end{aligned}$

and

$\begin{aligned} J_n^2(z) = \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{-t} \cos{zt} \, dt . \end{aligned}$

It follows that

$\begin{aligned} \frac{d}{dz} J_n^1(z) & = \int_{0}^{\infty} t^{\frac{1}{n}} e^{-t} \cos{zt} \, dt \\ & = \left[ \frac{t^{\frac{1}{n}} e^{-t}}{1 + z^2} (-\cos{zt} + z\sin{zt}) \right]_{0}^{\infty} - \frac{1}{n(1+z^2)} \int_0^{\infty} t^{\frac{1}{n}-1} e^{-t} (-\cos{zt} + z \sin{zt}) \, dt \, \text{ , (Integrate by part with} \, u = t^{1/n}, dv = e^{-t} \cos{zt} dt \text{)} \\ & = - \frac{z}{n(1+z^2)}J_n^1 + \frac{1}{n(1+z^2)}J_n^2 \end{aligned}$

and

$\begin{aligned} \frac{d}{dz} J_n^2(z) & = -\int_{0}^{\infty} t^{\frac{1}{n}} e^{-t} \sin{zt} \, dt \\ & = \left[ -\frac{t^{\frac{1}{n}} e^{-t}}{1 + z^2} (-\sin{zt} - z\cos{zt}) \right]_{0}^{\infty} + \frac{1}{n(1+z^2)} \int_0^{\infty} t^{\frac{1}{n}-1} e^{-t} (-\sin{zt} + z \cos{zt}) \, dt \, \text{ , (Integrate by part with} \, u = t^{1/n}, dv = e^{-t} \sin{zt} dt \text{)} \\ & = - \frac{1}{n(1+z^2)}J_n^1 - \frac{z}{n(1+z^2)}J_n^2 . \end{aligned}$

We can write this system of differential equations as

$\begin{aligned} \frac{d}{dt} \begin{bmatrix} J_n^1(z) \\ J_n^2(z) \end{bmatrix} = \begin{bmatrix} -\frac{z}{n(1+z^2)} & \frac{1}{n(1+z^2)} \\ -\frac{1}{n(1+z^2)} & -\frac{z}{n(1+z^2)} \end{bmatrix} \begin{bmatrix} J_n^1(z) \\ J_n^2(z) \end{bmatrix} \end{aligned}$

with initial condition

$\begin{aligned} \begin{bmatrix} J_n^1(0) \\ J_n^2(0) \end{bmatrix} = \begin{bmatrix} \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{-t} \sin{0t} \, dt \\ \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{-t} \cos{0t} \, dt \end{bmatrix} = \begin{bmatrix} 0 \\ \Gamma(\frac{1}{n}) \end{bmatrix} \end{aligned}$

Let

$\begin{aligned} A(z) = \begin{bmatrix} -\frac{z}{n(1+z^2)} & \frac{1}{n(1+z^2)} \\ -\frac{1}{n(1+z^2)} & -\frac{z}{n(1+z^2)} \end{bmatrix} . \end{aligned}$

Then,

$\begin{aligned} \int_{0}^{z} A(z) \, dz = \begin{bmatrix} -\frac{1}{2n} \log{(1+z^2)} & \frac{1}{n} \arctan{z} \\ -\frac{1}{n} \arctan{z} & -\frac{1}{2n} \log{(1+z^2)} \end{bmatrix} \end{aligned}$

and

$\begin{aligned} e^{\int_{0}^{z} A(z) \, dz} = \begin{bmatrix} (1+z^2)^{-\frac{1}{2n}} \cos{(\frac{\arctan{z}}{n})} & (1+z^2)^{-\frac{1}{2n}} \sin{(\frac{\arctan{z}}{n})} \\ -(1+z^2)^{-\frac{1}{2n}} \sin{(\frac{\arctan{z}}{n})} & (1+z^2)^{-\frac{1}{2n}} \cos{(\frac{\arctan{z}}{n})} \end{bmatrix} . \end{aligned}$

The solution of the differential equations is

$\begin{aligned} \begin{bmatrix} J_n^1(z) \\ J_n^2(z) \end{bmatrix} = e^{\int_{0}^{z} A(z) \, dz} \begin{bmatrix} J_n^1(0) \\ J_n^2(0) \end{bmatrix} = \begin{bmatrix} (1+z^2)^{-\frac{1}{2n}} \cos{(\frac{\arctan{z}}{n})} & (1+z^2)^{-\frac{1}{2n}} \sin{(\frac{\arctan{z}}{n})} \\ -(1+z^2)^{-\frac{1}{2n}} \sin{(\frac{\arctan{z}}{n})} & (1+z^2)^{-\frac{1}{2n}} \cos{(\frac{\arctan{z}}{n})} \end{bmatrix} \begin{bmatrix} 0 \\ \Gamma(\frac{1}{n}) \end{bmatrix} = \begin{bmatrix} (1+z^2)^{-\frac{1}{2n}} \sin{(\frac{\arctan{z}}{n})} \Gamma(\frac{1}{n}) \\ (1+z^2)^{-\frac{1}{2n}} \cos{(\frac{\arctan{z}}{n})} \Gamma(\frac{1}{n}) \end{bmatrix} \end{aligned}$

Finally, the given integral, using $n = 7$ , can be evaluated as

$\begin{aligned} I_7 & = \frac{1}{7} J_7^1(1) \\ & = \frac{\sin{(\frac{\pi}{28})} \Gamma(\frac{8}{7})}{\sqrt[14]{2}} \end{aligned}$