3 votes 3 votes $f(x) = \int_{- \infty}^{\infty} f(x) e^{2\pi x} dx$ solve f(x) Calculus integration definite-integral calculus engineering-mathematics + – Pavan Kumar Munnam asked Oct 11, 2017 Pavan Kumar Munnam 2.5k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply srestha commented Oct 11, 2017 reply Follow Share I think unsolvable as derivation power keep on incrementing 0 votes 0 votes Pavan Kumar Munnam commented Oct 11, 2017 reply Follow Share F(x) can be 0 also? 0 votes 0 votes srestha commented Oct 11, 2017 reply Follow Share No if u put 0 e2pix becomes 1 but it cannot be only 0 a range should be there 0 votes 0 votes Pavan Kumar Munnam commented Oct 11, 2017 reply Follow Share But it should be equal to f(x) also only 0 can satisfy it 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes Someone verify if it is correct! EDIT : Best answer is in the comments just_bhavana answered Oct 11, 2017 edited Oct 12, 2017 by just_bhavana just_bhavana comment Share Follow See all 2 Comments See all 2 2 Comments reply joshi_nitish commented Oct 12, 2017 reply Follow Share @just_bhavana you have changed the actual qsn...answer will be like this, f(x) = $\int_{-\infty }^{\infty }f(x)e^{2\pi x}dx$ = k //since definite integral will be some constant. f(x) = k now, since we know that f(x)=k is some constant independent of x, new equation will be, f(x) = f(x)$\int_{-\infty }^{\infty }e^{2\pi x}dx$ //since f(x) is constant, it can be pulled out of integral sign. now, f(x) = 0 is clearly one solution, now, suppose, f(x) = k(k$\neq$0), then above equation will decompose into k = k$\int_{-\infty }^{\infty }e^{2\pi x}dx$ 1 = $\int_{-\infty }^{\infty }e^{2\pi x}dx$ // this is incorrect. so we can say that f(x) is constant function and f(x) = 0,$\forall$x$\epsilon (-\infty ,\infty )$ 1 votes 1 votes just_bhavana commented Oct 12, 2017 i edited by just_bhavana Oct 12, 2017 reply Follow Share yes, thank you for pointing out! even I was confused as how a constant value can be a function of x. Wonderful explanation :) 0 votes 0 votes Please log in or register to add a comment.