DIFFERENTIATING UNDER THE INTEGRAL SIGN

1 DIFFERENTIATING UNDER THE INTEGRAL SIGNKEITH CONRADI had learned to do integrals by various methods shown in a book that my highschool physics teacher Mr. Bader had given me. [It] showed how to differentiateparameters UNDER the INTEGRAL sign it s a certain operation. It turns out that snot taught very much in the universities; they don t emphasize it. But I caught onhow to use that method, and I used that one damn tool again and again. [If] guysat MIT or Princeton had trouble doing a certain INTEGRAL , [then] I come along andtry DIFFERENTIATING UNDER the INTEGRAL sign, and often it worked.

2 So I got a greatreputation for doing integrals, only because my box of tools was different fromeverybody else s, and they had tried all their tools on it before giving the problemto Feynman [5, pp. 71 72] method of differentiation UNDER the INTEGRAL sign, due to Leibniz in 1697 [4], concerns integralsdepending on a parameter, such as 10x2e txdx. Heretis the extra parameter. (Sincexis thevariable of integration,xisnota parameter.) In general, we might write such an INTEGRAL as( ) baf(x,t) dx,wheref(x,t) is a function of two variables likef(x,t) =x2e (x,t) = (2x+t3)2.

3 Then 10f(x,t) dx= 10(2x+t3)2dx. An anti-derivativeof (2x+t3)2with respect toxis16(2x+t3)3, so 10(2x+t3)2dx=(2x+t3)36 x=1x=0=(2 +t3)3 t96=43+ 2t3+ answer is a function oft, which makes sense since the integrand depends ont. We integrateoverxand are left with something that depends only ont, INTEGRAL like baf(x,t) dxis a function oft, so we can ask about itst-derivative, assumingthatf(x,t) is nicely behaved. The rule, calleddifferentiation UNDER the INTEGRAL sign, is that thet-derivative of the INTEGRAL off(x,t) is the INTEGRAL of thet-derivative off(x,t):( )ddt baf(x,t) dx= ba tf(x,t) similar story with integration by parts in the first before this quote, Feynman wrote One thing I never did learn was contour integration.

4 Perhaps he meantthat he never felt he learned itwell, since he did know it. See [6, Lect. 14, 15, 17, 19], [7, p. 92], and [8, pp. 47 49].A challenge he gave in [5, p. 176] suggests he didn t like contour CONRADIf you are used to thinking mostly about functions with one variable, not two, keep in mind that( ) involves integrals and derivatives with respect toseparatevariables: integration with respecttoxand differentiation with respect saw in Example that 10(2x+t3)2dx= 4/3 + 2t3+t6, whoset-derivative is6t2+ 6t5.

5 According to ( ), we can also compute thet-derivative of the INTEGRAL like this:ddt 10(2x+t3)2dx= 10 t(2x+t3)2dx= 102(2x+t3)(3t2) dx= 10(12t2x+ 6t5) dx= 6t2x2+ 6t5x x=1x=0= 6t2+ answer agrees with our first, more direct, will apply ( ) to many examples of integrals, in Section 12 we will discuss the justificationof this method in our examples, and then we ll give some more s factorial INTEGRAL in a new lightFor integersn 0, Euler s INTEGRAL formula forn! is( ) 0xne xdx=n!,which can be obtained by repeated integration by parts starting from the formula( ) 0e xdx= 1whenn= 0.

6 Now we are going to derive Euler s formula in another way, by repeated differentiationafter introducing a parametertinto ( ).Fort >0, letx=tu. Then dx=tduand ( ) becomes 0te tudu= bytand writinguasx(why is this not a problem?), we get( ) 0e txdx= is a parametric form of ( ), where both sides are now functions oft. We needt >0 in orderthate txis integrable over the regionx we bring in differentiation UNDER the INTEGRAL sign. Differentiate both sides of ( ) withrespect tot, using ( ) to treat the left side.

7 We obtain 0 xe txdx= 1t2, DIFFERENTIATING UNDER THE INTEGRAL SIGN3so( ) 0xe txdx= both sides of ( ) with respect tot, again using ( ) to handle the left side. We get 0 x2e txdx= out the sign on both sides,( ) 0x2e txdx= we continue to differentiate each new equation with respect tota few more times, we obtain 0x3e txdx=6t4, 0x4e txdx=24t5,and 0x5e txdx= you see the pattern? It is( ) 0xne txdx=n!tn+ have used the presence of the extra variabletto get these equations by repeatedly applyingd/dt.

8 Now specializetto 1 in ( ). We obtain 0xne xdx=n!,which is our old friend ( ). Voil a!The idea that made this work is introducing a parametert, using calculus ont, and then settingtto a particular value so it disappears from the final formula. In other words, sometimesto solvea problem it is useful to solve a more general problem. Compare ( ) to ( ). damped sine integralWe are going to use differentiation UNDER the INTEGRAL sign to prove 0e txsinxxdx= 2 arctantfort > this integralF(t) and setf(x,t) =e tx(sinx)/x, so ( / t)f(x,t) = e txsinx.

9 ThenF (t) = 0e tx(sinx) integrande txsinx, as a function ofx, can be integrated by parts: eaxsinxdx=(asinx cosx)1 + CONRADA pplying this witha= tand turning the indefinite INTEGRAL into a definite INTEGRAL ,F (t) = 0e tx(sinx) dx=(tsinx+ cosx)1 +t2e tx x= x= ,tsinx+ cosxoscillates a lot, but in a bounded way (since sinxand cosxare boundedfunctions), while the terme txdecays exponentially to 0 sincet >0. So the value atx= is (t) = 0e tx(sinx) dx= 11 + explicit antiderivative of 1/(1 +t2), namely arctant.

10 SinceF(t) has the samet-derivative as arctant, they differ by a constant: for some numberC,( ) 0e txsinxxdx= arctant+Cfort > ve computed the INTEGRAL , up to an additive constant, without finding an antiderivative ofe tx(sinx) computeCin ( ), lett on both sides. Since|(sinx)/x| 1, the absolute value ofthe INTEGRAL on the left is bounded from above by 0e txdx= 1/t, so the INTEGRAL on the left in( ) tends to 0 ast . Since arctant /2 ast , equation ( ) ast becomes0 = 2+C, soC= /2.