350 THEORY OF HEAT. [CHAP. IX. 



the integral sign in order that the expression I dqQe~ qx may be 



equal to a given function, the integral being taken from q nothing 

 to q infinite 1 ? But without stopping for different consequences, 

 the examination of which would remove us from our chief object, 

 we shall limit ourselves to the following result, which is obtained 

 by combining the two equations (e) and (e). 



They may be put under the form 



- 7rf(x) = I dq sin qx I dzf (a) sin qx, 



A * Jo Jo 



1 /-co roo 



and ~ TrF (x) = I dq cos qx daF (a) cos qx. 



* Jo &quot;Jo 



If we took the integrals with respect to a. from oo to -f oo, 

 the result of each integration would be doubled, which is a neces 

 sary consequence of the two conditions 



/() = -/(-) and F(*)=F (-a). 

 We have therefore the two equations 



-CO ,00 



7rf(x) = I dq sin qx I dxf(&amp;lt;x) sin qx, 

 Jo J- 



,00 -00 



and TrF (x) = I dq cos qx I r/aF(a) cos qx. 



JO J-oo 



We have remarked previously that any function $ (x) can 

 always be decomposed into two others, one of which F (x) satisfies 

 the condition F(x) F(x], and the other f(x) satisfies the 

 condition /(#) = /( x). We have thus the two equations 



/+oo /H-oo 



dzF (a) sin ^a, and = I dxf(oL) cos qx, 

 -oo J -oo 



1 To do this write x*J - 1 in f(x) and add, therefore 



2 JQ, cos qx dq =f (x J~l) +f(-x A^l), 

 which remains the same on writing - x for x, 



therefore Q = - jdx [f(x,J~l} +f(-x J^l)] cos qx dx. 



Again we may subtract and use the sine hut the difficulty of dealing with 

 imaginary quantities recurs continually. [R. L. E.] 



