This page lists a variety of integrals that are highly useful in mathematical physics, along with their derivations. The latter are revealed by clicking the appropriate button.

Contents

Binomial Integrals
Trigonometric Integrals
Exponential Integrals
Gaussian Integrals
Miscellaneous Integrals


Binomial Integrals

dx(1 -axm )r =x F12 (r, 1m; 1m+1; axm)

Given the general binomial expansion

(1-x )r =1-rx +r (r-1) 2! x2 -r(r-1) (r-2) 3! x3 + (1-x )r =k=0 Γ(r +k) Γ(r) xkk!

the evaluation of the integral is

dx (1-axm )r =dx k=0 Γ(r +k) Γ(r) ak k! xmk dx (1-axm )r =k=0 Γ(r +k) Γ(r) ak k! xmk+1 mk+1 dx (1-axm )r =k=0 Γ(r +k) Γ(r) 1m 1m+k ak k! xmk+1 dx (1-axm )r =k=0 Γ(r +k) Γ(r) Γ(1m +k) Γ(1m) Γ(1m +1) Γ(1m +1+k) ak k! xmk+1 dx (1-axm )r =x F12 (r, 1m; 1m+1; axm)

using Γ(x+1) =xΓ(x) in the fourth step. The function in the result is the Gauss hypergeometric function. The result is a finite polynomial if r is a positive integer.



dx xp(1 -axm )r =xp+1 p+1 F12 (r, p+1m ; p+1m +1; axm)

Change the variable of integration and use the previous result:

dxxp (1-axm )r =1p+1 d( xp+1 ) [1-a(x p+1 )m/(p +1) ]r dxxp (1-axm )r =xp+1 p+1 F12 (r, p+1m ; p+1m +1; axm)



dx( xp -axm )r =xpr +1 pr+1 F12 (r, pr+1 m-p ; pr+1 m-p +1; axm-p)

Factor out the first term inside the binomial and use the previous result:

dx( xp -axm )r =dx xpr (1-ax m-p )r dx( xp -axm )r =xpr +1 pr+1 F12 (r, pr+1 m-p ; pr+1 m-p +1; axm-p)



dx xp (1-a1 xm )r1 (1-an xm )rn =xp+1 p+1 FD(n)( p+1m; r1, , rn; p+1m +1; a1xm, , anxm)

Given the general binomial expansion

(1-x )r =1-rx +r (r-1) 2! x2 -r(r-1) (r-2) 3! x3 + (1-x )r =k=0 Γ(r +k) Γ(r) xkk!

the evaluation of the integral is

dx xp (1-a1 xm )r1 (1-an xm )rn =dx k1,, kn=0 Γ(r1 +k1) Γ(r1 ) Γ(rn +kn) Γ(rn ) a1k1 k1! ankn kn! xmk1+ +mkn +p =xp+1 k1,, kn=0 Γ(r1 +k1) Γ(r1 ) Γ(rn +kn) Γ(rn ) 1m k1+ +kn +p+1 m ×(a1xm )k1 k1! (anxm )kn kn! =xp+1 p+1 k1,, kn=0 Γ(r1 +k1) Γ(r1 ) Γ(rn +kn) Γ(rn ) ×Γ(p+1 m+k1+ +kn) Γ(p+1 m) Γ(p+1 m+1) Γ(p+1 m+1+k1+ +kn) ×(a1xm )k1 k1! (anxm )kn kn! =xp+1 p+1 FD(n)( p+1m; r1, , rn; p+1m +1; a1xm, , anxm)

using Γ(x+1) =xΓ(x) in the third step. The function in the result is the n-dimensional Lauricella hypergeometric function of n variables. The exponent m must be the same in all binomials in order to factor it out from the other parameters in the second step.



dx xp (1-x)q (1-a1x )r1 (1-anx )rn =B(p+1, q+1) xp+1 FD(n)( p+1; r1, , rn; p+q+2; a1x, , anx)

This is a special case of the previous result with summation over one index. Begin as above but stop at the second step:

dx xp (1-x)q (1-a1x )r1 (1-anx )rn =dx k,k1,, kn=0 Γ(r1 +k1) Γ(r1 ) Γ(rn +kn) Γ(rn ) a1k1 k1! ankn kn! ×1k! Γ(q +k) Γ(q) xk+k1+ +kn +p =xp+1 k,k1,, kn=0 Γ(r1 +k1) Γ(r1 ) Γ(rn +kn) Γ(rn ) (a1x )k1 k1! (anx )kn kn! ×1k! Γ(q +k) Γ(q) 1 k+k1+ +kn +p+1

Complete the summation over k

k=0 1k! Γ(q +k) Γ(q) 1 k+k1+ +kn +p+1 =k=0 1k! Γ(q +k) Γ(q) Γ(k +k1+ +kn +p+1) Γ(k +k1+ +kn +p+2) =Γ( k1+ +kn +p+1) Γ( k1+ +kn +p+2) ×F12 (q, k1+ +kn +p+1; k1+ +kn +p+2;1) =Γ( k1+ +kn +p+1) Γ(q+1) Γ( k1+ +kn +p+q+2)) Γ(1)

using Γ(x+1) =xΓ(x) in the first step and the Gauss summation formula

F12 (a,b; c,1) =Γ(c) Γ(c-a-b) Γ(c-a) Γ(c-b)

in the third. The final integral is

dx xp (1-x)q (1-a1x )r1 (1-anx )rn =Γ(p+1) Γ(q+1) Γ(p+q +2) xp+1 FD(n)( p+1; r1, , rn; p+q+2; a1x, , anx) =B(p+1, q+1) xp+1 FD(n)( p+1; r1, , rn; p+q+2; a1x, , anx)

It is identical to setting q=0 and m=1 in the previous result.



01 dx xa-1 (1-x) c-a-1 (1-z1x )b1 (1-z2x )b2 (1-z3x )b3 =B(a, c-a) FD( a; b1, b2, b3; c; z1, z2, z3)

This is a special case of the previous result. It is the Euler-type integral representation of a Lauricella triple hypergeometric function.



01 dx xa-1 (1-x) c-a-1 (1-z1x )b1 (1-z2x )b2 =B(a, c-a) F1( a; b1, b2; c; z1, z2)

This is a special case of the previous result. It is the Euler-type integral representation of an Appell double hypergeometric function.



01 dx xa-1 (1-x) c-a-1 (1-zx )b =B(a, c-a) F12 (a, b; c; z)

This is a special case of the previous result. It is the Euler-type integral representation of the Gauss hypergeometric function. Since this function is symmetric in its first two parameters, they can be interchanged on both sides of the relation.




Trigonometric Integrals

dθ sinpθ cosmθ =sin p+1θ p+1 F12 (1-m2 , p+12 ; p+32 ; sin2θ)

Change variable to x=sinθ and use a binomial integral:

dθ sinpθ cosmθ =dx xp(1 -x2 )(m-1) /2 dθ sinpθ cosmθ =sin p+1θ p+1 F12 (1-m2 , p+12 ; p+32 ; sin2θ)



dθ sinpθ cosmθ =cos m+1θ m+1 F12 (1-p2 , m+12 ; m+32 ; cos2θ)

Change variable to x=cosθ and proceed as before: sine is replaced with cosine, exponents are interchanged and the result has an additonal overall minus sign.



0 π/2dθ sinpθ cosmθ =Γ(p +12) Γ(m +12) 2Γ(p +m2 +1)

Given the relationship of the beta function to the gamma function and the former’s trigonometric representation,

B(x,y) =Γ(x) Γ(y) Γ(x+y) =20 π/2 dθ sin2x-1θ cos2y-1θ

the integral is trivial. Alternately given the Gauss summation formula

F12 (a,b; c,1) =Γ(c) Γ(c-a-b) Γ(c-a) Γ(c-b)

it can be evaluated from either of the previous two results:

0 π/2dθ sinpθ cosmθ =1p+1 F12 (1-m2 , p+12 ; p+32 ; 1) =Γ(p +32) Γ(m +12) (p+1) Γ(p+m2 +1) Γ(1) =Γ(p +12) Γ(m +12) 2Γ(p +m2 +1)

0 π/2dθ sinpθ cosmθ =1m+1 F12 (1-p2 , m+12 ; m+32 ; 1) =Γ(m +32) Γ(p +12) (m+1) Γ(m+p2 +1) Γ(1) =Γ(p +12) Γ(m +12) 2Γ(p +m2 +1)




Exponential Integrals

dx xn ex =n! ex k=0 n xkk! n! ex en(x)

Integrate by parts n times:

dx xn ex =xn ex +ndx xn-1 ex dx xn ex =[xn +nxn-1 ]ex +n(n-1) dx xn-2 ex dx xn ex =[xn +n! (n-1)! xn-1 + +n! 1!x] ex +n! dx ex dx xn ex =n! ex k=0 n xkk!

From the definition of the incomplete gamma function, this can also be written

dx xn ex =Γ( n+1,x)

since the exponential vanishes at the upper limit.



dx xn ex =(1 )nn! ex en(x)

Substitute x for x in the previous result.



dx xn eax =(1 )n n! an+1 eax en (ax)

Substitute ax for x in the previous result.




Gaussian Integrals

dx eax2 =πa

Square the original integral and change to polar coordinates:

dx eax2 =[ dx eax2 × dy eay2 ]12 dx ea x2 =[0 rdr 02π dφ ear2 ]12 dx ea x2 =[π 0 dr2 ear2 ]12 dx eax2 =πa



dx eiax2 =iπa

Substitute ia for a in the previous result.



dx eax2 +bx= πa exp(b2 4a)

Complete the square in the exponential and use a previous result:

dx eax2 +bx = dx exp[a(x -b2a )2 +b2 4a] dx eax2 +bx =exp(b2 4a) dy ea y2 dx eax2 +bx= πa exp(b2 4a)




Miscellaneous Integrals

cx dt ct dt ct y(t)dt =1 (n-1)! cx (x-t )n-1 y(t)dt

This equivalence can be established directly with the Leibniz integral rule, which is

ddx a(x) b(x) f(x,t) dt =f[x, b(x)] db(x) dx -f[x, a(x)] da(x) dx + a(x) b(x) f(x,t) xdt

When applied to the equivalence, the second term of this expression is zero on both sides due to the constant lower limit of integration. On the left-hand side of the quivalence, the third term is zero and the first merely removes one instance of integration:

ddx cx dt ct dt ct y(t)dt =cx dt ct y(t)dt

On the right-hand side of the equivalence, the first term is zero because the integrand is zero at the upper limit of integration. The partial differentiation in the third term gives

1 (n-1)! ddx cx (x-t )n-1 y(t)dt =1 (n-2)! cx (x-t )n-2 y(t)dt

which is precisely what is expected for one less instance of integration.




Uploaded 2020.04.21 — Updated 2020.07.22 analyticphysics.com