which is the same result as (77), Art. 94. Ex. 4. By means of the preceding integral may another integral be determined which expresses a remarkable discontinuous function, of which considerable use will be made in the sequel. Since 2 sin ma cos nx = sin (m+n) x + sin (m—n)x, Hence, if positive values only of m and n are taken, tan -1 ; a (110) when n = m; and = 0, for all values of n greater than m. :) dx = a πλα π b b (a + b) ra da 0 a + b π [log (+) = log. (111) 2 (ay) da fo dx. -ax dx = ; dy e-a da ; 0 2 (ay) Now, since a and y are two quantities independent of each other, and as each, being a definite integral, is a constant, the repetition is equivalent to multiplication; and we have in the right-hand member replace y by ax2; then, since the left an integral which we shall hereafter find of considerable importance. If a = 1, 101.] When a definite integral, which is to be evaluated, is differentiated or integrated with respect to a variable parameter contained in the element-function, another definite integral is produced, which can frequently be evaluated by means of its indefinite integral; in other cases the new integral can often be integrated either by parts or by some other method: and sometimes the original definite integral will arise in the process. When this is the case, a differential equation is formed, the solution of which will give the value of the definite integral. Sometimes again the definite integral will arise, when two or more differentiations or integrations have been effected on the original integral. The following examples illustrate the method, and the development of it will be best understood by them. As the solution is definite, the integrals of both sides of this equation must be definite. Let the limits of r ber and 0; and |