Ytukyg

The integral is,
$$I=int_{-infty}^{infty}frac{sin x}{x}, dx$$
I know the answer would be $pi$ and I know how to solve this using Feynman's method and Fourier transform. However I was trying something, rather naively, since by understanding of the subject is constrained to undergraduate mathematical physics. I did the following, taking the imaginary part of $e^{ix}$,
$$int_{-infty}^{infty}frac{sin x}{x}, dx=Imleft{int_{-infty}^{infty}frac{e^{ix}}{x}, dxright}$$
Then I introduced a variable $alpha$ where taking $alpha=1$ would result in the required integral. So we have here,
$$I(alpha)=Imleft{int_{-infty}^{infty}frac{e^{ialpha x}}{x}, dxright}$$
Taking the derivative of $I$ with $alpha$ we will get,
$$frac{dI(alpha)}{dalpha}=Imleft{int_{-infty}^{infty}ie^{ialpha x}, dxright}$$
We know that this is the definition of Dirac delta, $delta(alpha)$. Therefore,
$$frac{dI(alpha)}{dalpha}=Imleft{i2pidelta(alpha)right}=2pidelta(alpha)$$
Integrating with $alpha$,
$$I(alpha)=2pitheta(alpha)+c$$
Where, $theta$ is the Heaviside step function and $c$ is the constant of integration. Using $alpha=-infty$ from the initial equation we shall get $I(-infty)=0$, which gives $c=0$. Therefore,
$$I(alpha)=2pitheta(alpha)$$
And putting $alpha=1$ here, we have,
$$I=2pi$$
Which isn't the required answer. I want to know where I am going wrong, which step, or if it's the entire method. I am also not aware of all the properties of the functions I used and this is just me fiddling with things, so I will be really obliged if anyone can help.

Edit for duplication claim: My question isn't exactly about how to solve the sinc integral using the known methods. I want to know what is wrong with the approach I am using, which I know is wrong, but I don't know what exactly is going wrong.

Edit after suggestion in comments: Okay assuming $alpha=-infty$ to calculate the integration constant's value, is kind of a blunder, since that's not gonna work. However, if I take $alpha=-1$ this might work, and will give me,
$$I(-1)=-int_{-infty}^{infty}frac{sin x}{x}, dx=-I(1)$$
This means,
$$c=I(-1)=-I(1)$$
And that would give me,
$$I(alpha)=2pitheta(alpha)-I(1)$$
And now taking $alpha=1$,
$$I=pi$$
And that matches the answer, however I am kinda skeptical if this approach is right.

So effectively your start-at-$-1$ approach boils down to $$I=frac{I-(-I)}{2}=frac{1}{2}int_{-1}^1 dalphafrac{d}{dalpha}Imint_{Bbb R}frac{e^{ialpha x} dx}{x}=int_{-1}^1 dalpha pidelta (alpha)=pi,$$which is indeed correct. In fact, the use of $[-1,,1]$ as our integration range can be enormously generalised. Note that susbtitution obtains $int_{Bbb R}frac{sin alpha x}{x}dx=Ioperatorname{sgn}alpha$ for any $alphainBbb{R}$, including $alpha =0$ viz. $operatorname{sgn}0=0$. Thus if we chose any values $a,,b$ of distinct sign we'd have $$I=frac{1}{operatorname{sgn}b-operatorname{sgn}a}int_a^b dalpha 2pidelta(alpha),$$so $I=pi$ follows provided it makes sense to say $$int_a^b dalpha 2pidelta(alpha)=pi(operatorname{sgn}b-operatorname{sgn}a).$$The only special case where you might question this is when $a$ or $b$ is $0$, whence we need $$b>0impliesint_0^{b}delta(alpha) dalpha=tfrac{1}{2}.$$But this is indeed a sensible characterisation of $delta (x)$ because $xmapsto -x$ implies $$int_0^{b}delta(alpha) dalpha=tfrac{1}{2}int_{-b}^{b}delta(alpha) dalpha=tfrac{1}{2}.$$Given the need to define $theta (x):=int_{-infty}^xdelta(alpha) dalpha$, we end up saying $theta(0)=tfrac{1}{2}$, making $theta$ neither left- nor right-continuous at $0$. Other conventions are sometimes used, in which case greater case is needed in this problem.