i!

Define the Factorial of a Complex number.

In usual sense, factorial is defined as,
$$n! = \prod_{k=1}^{n} k=1 \cdot2\cdot3\ldots(n-2)\cdot(n-1)\cdot n \tag*{(1)}$$
Now, the not so usual definition is based on the famous Gamma Function,
$$\Gamma(t) = \int_0^\infty e^{-x}x^{t-1}\mathrm{d}x \tag*{(2)}$$
There is an unique and very useful property,
$$\Gamma(n+1)=n! \tag*{(3)}$$
To extend $(3)$ into the complex domain, we will first have to go through Analytic Continuation, please read about Analytic Continuation here.
Therefore, after analytic continuation, we can write it as,
$$z! \overset{\text{def}}{=} \Gamma(z+1) \tag*{(4)}$$
For, $z \in \mathcal C-{-1,-2...}$
So, now, clearly,
$$i!= \Gamma(i+1) \tag*{(5)}$$
By $(3)$
$$\Gamma(i+1)=\int_0^\infty e^{-x} x^{(i+1)-1}\mathrm{d}x \tag*{(6)}$$
Clearly,
$$\Gamma(i+1)=\int_0^\infty e^{-x} x^{i}\mathrm{d}x \tag*{(6)}$$
For easier computation, please catch that,
$$x^i = e^{i \ln x} = \cos ( \ln x)+i \sin( \ln x)$$
Let’s break it down,
$$i! = \int_0^\infty e^{-x} (\cos ( \ln x)+i \sin( \ln x))\mathrm{d}x$$
$$\implies i! = \int_0^\infty e^{-x} \cos ( \ln x)\mathrm{d}x+i\int_0^\infty e^{-x} \sin ( \ln x)\mathrm{d}x$$
If you have reached this far, you obviously know how to solve the above integral.
Cheers!

Integration : Problem 2

$$\text{I have been inactive for quite a long time.}\\ \text{As usual, I will start this blogging streak}\\ \text{with a small warm-up exercise.}\\ \text{I will solve some mildly complex Integrals, }$$

$\underline{\text{Question 1.}} \\$ $\LARGE \int _{0}^{\frac{1}{2}}\frac{x^2}{(1+x^2)^{3/2}} \mathrm{d}x$
$\underline{\text{Solution}}$:
$$\text{Let, } I_{1} = \int _{0}^{\frac{1}{2}}\frac{x^2}{(1+x^2)^{3/2}} \mathrm{d}x \\ \text{We can write, } I_1 = \int _{0}^{\frac{1}{2}} \frac{x^2 + 1 - 1}{(1+x^2)^{3/2}} \mathrm{d}x \\ \implies I_1 = \int _{0}^{\frac{1}{2}} \frac{x^2+1}{(1+x^2)^{3/2}}\mathrm{d}x - \int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{Which with some algebraic manipulation comes to, }\\ I_1 = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1+x^2}}\mathrm{d}x - \int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{Let,} I_2 = \int \frac{1}{\sqrt{1+x^2}}\mathrm{d}x \\ \text{and, Let , } I_3 = \int \frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{We know that, } \boxed{\boxed{\int \frac{\mathrm{d}x}{\sqrt{x^2 + a^2}} = \log|x +\sqrt{x^2 + a^2}|+c }} \\ \therefore I_2 = \int \frac{1}{\sqrt{1+x^2}}\mathrm{d}x = \log|x+\sqrt{1+x^2}| \\ \text{ neglecting } c \text{ because it will ultimately} \\ \text{ be a definite integral }.\\ \\ \text{Now the rigor starts, } I_3 = \int \frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{Take, } x = \tan {t} \rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = \sec^2 {t} \\ \therefore I_3 = \int \frac{\sec^2{t}}{(1+\tan^2{t})^{3/2}} \mathrm{d}t\\ \text{We know, } \boxed{\boxed{1 + \tan^2 \theta = \sec^2 \theta}} \\ \text{Using this } I_3 \text{ changes to, }\\ I_3 = \int \frac{1}{\sec t} \mathrm{d} t \implies \int \cos {t} \mathrm{d}t = \sin t \\ \text{We get the substitution in place ,}\\ t = \tan^{-1} x \rightarrow \sin t \rightarrow \sin{(\tan^{-1} x)} = \frac{x}{\sqrt{x^2+1}} \\ \therefore I_3 = \sin t = \frac{x}{\sqrt{x^2+1}} \text{Now,} \\ I_1 = I_2 - I_3 \text{ with no limits. }\\ \therefore I_1 = \int \frac{1}{\sqrt{1+x^2}}\mathrm{d}x = \log|x+\sqrt{1+x^2}| - \frac{x}{\sqrt{x^2+1}}\\ \text{Put limits } \frac{1}{2} \text{ to } 0 \text{ and get answer as, }\\ I_1 = {I_1}|_{0}^{1/2}=\boxed{0.033}.\\ \text{Cheers! } \bowtie$$

Integration : More Problems

Problem 2: $\int \frac{\ln(x)}{x} \ dx$

Solution :
Substitute,
$y$ = $\ln(x)$
$\implies dy$ = $\large \frac{dx}{x}$
Thus,
$I$ = $\int \frac{\ln(x)}{x}dx$
$\implies I$ = $\int y dy$
$\implies I$ = $\frac{y^{2}}{2} dy$
$\implies I$ = $\frac{(\ln(x))^{2}}{2} + C$

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Problem 3 : $\int x^{2} \sqrt{1-x^{2}}\ dx$

Solution:
Substitute,
$x = \sin(t)$
$\implies dx = \cos(t) \ dt$
Thus, the integral beomes,
$I$ = $\int \sin^{2}t \sqrt{1-sin^{2}t} \cos t \ dt$
$I$ = $\int \sin^{2}t \cos t \cos t \ dt$
$I$ = $\int \sin^{2}t \cos^{2} t \ dt$

Recall:
$\sin(2\theta) = 2 \ \sin \theta \ \cos \theta$
$\implies \sin \theta \ \cos \theta = \frac{\sin(2\theta)}{2}$

$I$ = $\int (\sin t \cos t)^{2} \ dt$
$I$ = $\int (\frac{\sin(2t)}{2})^{2} \ dt$
$I$ = $\frac{1}{4}\int \sin^{2}(2t) \ dt$

Recall:
$\cos (2 \theta) = \cos^{2} \theta - \sin^{2} \theta$
$\cos (2 \theta) = 1 - 2\sin^{2} \theta$
$\sin^{2} \theta = \frac{1 - \cos (2 \theta)}{2}$

$I$ = $\frac{1}{8}\int (1-\cos(4t)) \ dt$
$I$ = $\frac{1}{8}[t-\frac{\sin(4t)}{4}$
$I$ = $\frac{1}{8}[arcsin(x)-x(1-2x^{2}) \sqrt(1-x^{2}) + c$ [After simplifying $\sin(4t)$ using the above two Recall formulae

Integration : Problem 1

Problem : $\large \int (\sqrt{x^{2} - 4})^{3} dx$

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Solution:

For this problems, we use Euler Substitutions,
Substitute, $\sqrt{x^{2} - 4}$ = $t-x$
$\implies \sqrt{x^{2} - 4}+x$ = $t$ —(1)
$\implies {x^{2} - 4}$ = $(t-x)^{2}$
$\implies {x^{2} - 4}$ = $t^{2}-2tx+ x^{2}$
$\implies {- 4}$ = $t^{2}-2tx$
$\implies {x}$ = $\frac{t^{2}+4}{2t}$ —(2)
$\large \implies \sqrt{x^{2} - 4}$ = $\large \frac{t^{2}+4}{2t}$ —(3)
From (2) and using the quotient rule of differentiation,
$\large dx$ = $\large \frac{t^{2}-4}{2t^{2}}dt$
Plugging these in the given problem we get,
$I$ = $\int (\sqrt{x^{2} - 4})^{3} dx$
$I$ = $\int (\frac{t^{2}+4}{2t})^{3} . \frac{t^{2}-4}{2t^{2}}dt$
$I$ = $\int \frac{(t^{2})^{3}+4^{3}+3.(t^{2})^{2}.4+3.(t^{2}).4^{2}}{8t^{3}} . \frac{t^{2}-4}{2t^{2}}dt$
$I$ = $\int \frac {t^{8}-16t^{6} + 96t^{4}-256t^{2}+256}{16t^{5}}$
$I$ = $\int [\frac{t^{3}}{16} - t + \frac{6}{t} - \frac{16}{t^{3}} + \frac{16}{t^{5}}]$$dt$
$I$ = $\frac{t^{4}}{64} - \frac{t^{2}} {2}+ 6\ln|t| + \frac{8}{t^{2}} -\frac{4}{t^{4}} + C$
Put value of $t$ from equation (1) and get the answer.
Regards.