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If $x^{-1}(t)$ and $ y^{-1}(t)$ denote the integrals of x(t) and y(t) defined by

$x^{-1}(t)=\displaystyle\int_{-\infty}^{t} x(\lambda)d\lambda$

$y^{-1}(t)=\displaystyle\int_{-\infty}^{t}y(\lambda)d\lambda$

then,

$(x*y)^{-1}$ which denotes the inverse integral of (x*y )is equal to:

(a) $x^{-1}*y^{-1}$

(b) $x*y$

(c)$x^{-1}*y$

(d)$x*y^{-1}$

i tried it follows:

from both conditions above we can easily infer following

$\dfrac{d}{dt}x^{-1}(t)=x(t)\implies sX^{-1}(s)=X(s)$

$\dfrac{d}{dt}y^{-1}(t)=y(t)\implies sY^{-1}(s)=Y(s)$

then i multiplied both(keeping in mind multiplication in freq. domain corresponds to convolution in time domain)

$s^2[X^{-1}(s)Y^{-1}(s)]=X(s).Y(s)$

$s[X^{-1}(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[sX^{-1}(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[X(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}] \implies x*y^{-1}=(x*y)^{-1}$

and also,

if i would've written as

$s[X^{-1}(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[X^{-1}(s)sY^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[X^{-1}(s)Y(s)]=\dfrac{X(s).Y(s)}{s}\implies x^{-1}*y=(x*y)^{-1}$

so, options (c,d) both should be correct but in the book the answer is given as

only option (c)

so , where i am doing mistake, and what should be the correct answer, i'm stuck

with this book's faulty answers...so,please help

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    $\begingroup$ Your notation is a bit confusing. Normally I would interpret $x^{-1}$ as $1/x$. And you also mention "inverse integral", what is that? But more importantly, you define $x^{-1}(t)$ as the integral over the real line, so the result is constant, and not a function of $t$. This is probably not what you mean. $\endgroup$ – Matt L. Jan 27 '18 at 15:37
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    $\begingroup$ @ Matt L ....nothing is specially mentioned about 'inverse' in the question i've written question as it is given in the book........i took 'inverse' .meaning .....'as integrator' due to division by 's(jw)' $\endgroup$ – Faraday Pathak Jan 27 '18 at 15:45
  • $\begingroup$ After the two first formulas you write "... which denotes the inverse integral ...". And the title of you question is "inverse convolution integral". And still, your first two integrals are just numbers, not functions of $t$. I think you're missing a $t$ as upper integration limit, don't you? $\endgroup$ – Matt L. Jan 27 '18 at 15:56
  • $\begingroup$ @Matt L....... in book upper limit is 'infinity' but should be 't'(by definition of integrator in time domain) ....you're right ....now what should be done ahead $\endgroup$ – Faraday Pathak Jan 27 '18 at 16:06
  • $\begingroup$ @ Matt L .....(x * y) denotes convolution so, i thought (x*y)^{-1} would denote 'inverse' of convolution as per definition of 'inverse' for this specific question........ $\endgroup$ – Faraday Pathak Jan 27 '18 at 16:09
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The integral of a signal $x(t)$

$$f(t)=\int_{-\infty}^tx(\tau)d\tau$$

can be written as the convolution of $x(t)$ with the unit step function $u(t)$:

$$f(t)=(x\star u)(t)$$

Consequently, the integral of a convolution

$$g(t)=\int_{-\infty}^t(x\star y)(\tau)d\tau$$

can be written as

$$g(t)=((x\star y)\star u)(t)$$

And due to associativity of convolution we have

$$g(t)=((x\star u)\star y)(t)=((y\star u)\star x)(t)$$

So options $(c)$ and $(d)$ in your question are both correct.

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  • $\begingroup$ @ Matt L.......thank you so much......you always exploit basic properties of convolution(mostly assosciativity) . ....your solutions are always nice..... $\endgroup$ – Faraday Pathak Jan 27 '18 at 17:08

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