Page 1 :
_ Differentiation Under, , Integral Sign, , , , , , , , , , Differentiation Under Integral sign, , , , , , , , SALE lel ty Brey s) itr ctartepty, Under Integral sign, , 11.1_ Introduction, , , , This is an advanced technique for evaluating certain, , types of definite integrals which are not easily integrable, otherwise., , A, © log (1 +cos a cos x), eg.1@) = f cos x ce, , 0, The value of this integral is a function of a (and not, of x), hence we denote it as I (ct), a is called as parameter., As the title indicates we shall be differentiating the, integrand with respect to the parameter., , These definite integrals depend upon one or more, parameters., , First we establish a theorem., Theorem :, b, , J £0, 0)- dx, a, , Let, I(@) =, , where a and b are independent of a then, , i) 1 ds :, —= fanaa ffm dx, , d, b, [Xrae ]o, a, , da, , , , , , The letter 0 to be pronounced as dabba., , Proof : We have, by definition :, at lim V(a+Sc)-1 (a), da ” 50-50 eenooee PD), b, “2 Ma+8o)= J 1 (%, a+ Bay) - dx, a, dl lim 4] fe, Wee ba Osa. Titers dx- Tifa) ax, , lim y[iensatta a+8a)-1(x, a, , * Sa 70, io, i im [ ee+seh—tho f Na, a0, a, aa va jim, ; da 0 2° ,da0, , - It], , 11.2 . Steps to be followed to Evaluate the, Integral, , Step (1): First denote the integral as I (a) for I (a)) etc., , Then by applying D.U.LS. find V(a), and, , evaluate it., Step (I): Integrate 1 (cz) with respect to a. It involves a, constant A., Step (III): A constant ‘A’ can be evaluated by assigning, * proper value to the parameter., , , , , , \