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Please help me sort this issue out.

The integration property in Fourier series is as follows:

So, for proving the above property, i took this approach:

This is where my doubt is. Some books and websites just put upper limit (ignoring the lower limit) and compare with (1) to conclude that

However, when the lower integral limit is substituted, we get

This value cannot be evaluated. Though some websites say that this corresponds to the "initial value"(?) and that is zero... etc., how does one justify this mathematically?

Any help is appreciated. Thank you.

Note that the antiderivative of a function is only defined up to a constant. Furthermore, note that if you integrate a periodic function, the result is not necessarily periodic. Let

$$x(t)=sum_k=-infty^inftya_ke^jkomega_0ttag1$$

Now we integrate $(1)$ with an arbitrary lower integration limit $t_0$. I'll say more about that later.

$$beginaligny(t)&=int_t_0^tx(tau)dtau\&=sum_k=-infty^inftya_kint_t_0^te^jkomega_0taudtau\&=a_0(t-t_0)+sum_k=-infty\kneq 0^inftyfraca_kjkomega_0e^jkomega_0t-sum_k=-infty\kneq 0^inftyfraca_kjkomega_0e^jkomega_0t_0tag2endalign$$

Clearly, if $a_0neq 0$, $y(t)$ isn't periodic, so we can't hope to obtain a formula for its Fourier coefficients. So in order to end up with a periodic function $y(t)$, we require that $a_0=0$.

If $a_0=0$, we get from $(2)$

$$y(t)=sum_k=-infty^inftyb_ke^jkomega_0ttag3$$

with

$$b_k=begincasesdisplaystylefraca_kjkomega_0,&kneq 0\displaystyle-sum_l=-infty\lneq 0^inftyfraca_ljlomega_0e^jlomega_0t_0,&k=0endcasestag4$$

So in general the antiderivative $y(t)$ has a non-zero DC component, which depends on the choice of the lower integration limit $t_0$.

The antiderivative

$$tildey(t)=sum_k=-infty\kneq 0^inftyfraca_kjkomega_0e^jkomega_0ttag5$$

is the specific antiderivative of $x(t)$ that has a zero DC Fourier coefficient, and this is the one that is meant by the formula that is usually given in textbooks.

Consequently, the expression

$$y(t)=int_-infty^tx(tau)dtautag6$$

for the antiderivative of $x(t)$ with Fourier coefficients $b_k=a_k/(jkomega_0)$, $kneq 0$, and $b_0=0$,
is at least inaccurate; it is in fact a sloppy way of expressing the specific antiderivative $tildey(t)$ given by $(5)$, i.e., the one with a zero DC coefficient.

It appears to me that the "initial value" issue would affect only the DC value, $a_0$, of the Fourier series. In my opinion, the author should have left the integrals as indefinite integrals (which are the same as the "anti-derivative") or have expressed this relationship only in terms of differentiation, and not integration.

Actually, now that I think of it, the original $x(t)$ would have to have $a_0 = 0$ in order for the integration theorem to make any sense. And the resulting Fourier series can have any finite DC value you want. You would have to determine the resulting DC value by other means.

To do this theorem properly, you must first do it for the derivative of $x(t)$ first and then apply the results in reverse.

I dunno what book you're using but there are both some notational and even mathematical problems with the expression of this problem.

Oh, hell...

Let's say you have two periodic functions, $x(t)$ and $y(t)$ having exactly the same period (and fundamental frequency):

$$ x(t) triangleq sumlimits_k=-infty^infty a_k , e^j k omega_0 t $$

$$ y(t) triangleq sumlimits_k=-infty^infty b_k , e^j k omega_0 t $$

where the period common to both is $frac2 piomega_0$.

Suppose you were to differentiate $y(t)$:

$$ y'(t) = sumlimits_k=-infty^infty j k omega_0 b_k , e^j k omega_0 t $$

you can see right away that the DC term of $y'(t)$ is zero (as it should be):

$$ j k omega_0 b_k bigg|_k=0 = 0 $$

Now let's say that you set that differentiated periodic function to $x(t)$:

$$ x(t) = y'(t) $$

That means $y(t)$ is the indefinite integral of $x(t)$. Then you see that all of the coefficients of $y(t)$ are well-defined in terms of the coefficients of $x(t)$ except for the DC coefficient, $b_0$, which could be any finite value.

$$ a_k = j k omega_0 b_k $$

or

$$ b_k = fraca_kj k omega_0 qquad qquad forall k in mathbbZ ne 0 $$.

That is, in my opinion, the only legit way to look at the integration of a Fourier series.