SECT. I.] AN INVERSE PROBLEM. 337 



Developing the sign of the integral, we write as follows, the 

 equation from which the value of Q must be derived : 



F(x) = dq Q t cos qjc + dqQ z cos q z x + dqQ 3 cos q z x + &c. 



In order to make all the terms of the second member dis 

 appear, except one, multiply each side by dxcosrx, and then 

 integrate with respect to x from x = to x mr, where n is an 

 infinite number, and r represents a magnitude equal to any one 

 of q lf q z , q 3 , &c., or which is the same thing dq, 2dq, 3dq, &c. Let 

 q i be any value whatever of the variable q f and q^ another value, 

 namely, that which we have taken for r; we shall have r =jdq, 

 and q = idq. Consider then the infinite number n to express how 

 many times unit of length contains the element dq, so that we 



have n = -r- . Proceeding to the integration we find that the 

 dq 



value of the integral Idx cos qx cos rx is nothing, whenever r and 



q have different magnitudes ; but its value is ^ UTT, when q = r. 



This follows from the fact that integration eliminates from the 

 second member all the terms, except one ; namely, that which 

 contains qj or r. The function which affects the same term 

 is Qj, we have therefore 



dx F (x) cos qx = dq Q } ^ nir, 

 and substituting for ndq its value 1, we have 



cos qx. 



Q (*&amp;gt; 



We find then, in general, -^ = dxF(x)cosqx. Thus, to 



2 Jo 



determine the function Q which satisfies the proposed condition, 

 we must multiply the given function F(x) by dxcosqx, and in- 



2 

 tegrate from x nothing to x infinite, multiplying the result by - ; 



that is to say, from the equation F(x] = ldqf(q) cos qx, we deduce 



2 r 



f(q}=-ld,jcF(x)cosqx, the function F(f) representing the 

 F. ii. 22 



