HEAVISIDE'S EXPANSION THEOREM 



489 



^^1^ T?^. = c VC/^ Zt^^^"^11 + erf iqj^l^)-] 



p- -j- CO- n ^qn 



+ 





4^1/2g-l(W4) 



1 



4^l/2gl(>r/4) 



e*"'!! +erf (-a;i/V^'^%i/2,| 



- 4coW2!-uw4) ^-*"{1 + erf (- co^/^e-^^^^/^)}] 



2 



/^[-gi(a,i-W4) ei-f (co'/V"^/^^/^) 



_|_ g-l(a>«_W4) gj-f (^l/2g-i(W4)^)/2)] 



2Cc 



[sin (coO^(co'/¥/2V2/x) 



+ COs(co/)C(co'/2/l/2V2/x)] 



The last transformation is obtained by means of formulae 14 and 15. 

 Example 3: 

 Evaluate 



^ 1 



^ ~ p4 _ 3^2 _|_ 2 ■ 



This can be solved by the expansion theorem in the usual way. A 

 somewhat shorter method is to use the generalized theorem with the 

 operator q = p^. Then 



1 



= l+E 



1 



'/'(Sn, /) 



g2 — 3g + 2 '^ ' « 2qr? — 3g„ 

 qn = 1,2; \p{qn, t) = cosh pn^'H. 

 See equivalent No. 4. So 



3? = I + I cosh /V2 — cosh /. 



Example 4: 

 Evaluate 



sinh ^/j^/^ sinh &(7 & , ^^ sinh 6g„ . j,p. , ., ,imn 

 smh a^' '2 smh aq a aqn cosh aq-n 



— sinh aq = i sin mg. 



