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.

# 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 $\Gamma \left(x+1\right)=x\Gamma \left(x\right)$ 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 $\Gamma \left(x+1\right)=x\Gamma \left(x\right)$ 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 $\Gamma \left(x+1\right)=x\Gamma \left(x\right)$ 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 $x=sin\theta$ 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 $x=cos\theta$ 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,

$B(x,y) =Γ(x) Γ(y) Γ(x+y) =2∫0 π/2 dθ sin2x-1θ cos2y-1θ$

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

$∫dx xn e−x =−n! e−x ∑k=0 n xkk! ≡−n! e−x en(x)$

Integrate by parts n times:

$∫dx xn e−x =−xn e−x +n∫dx xn-1 e−x ∫dx xn e−x =−[xn +nxn-1 ]e−x +n(n-1) ∫dx xn-2 e−x ∫dx xn e−x =−[xn +n! (n-1)! xn-1 +⋯ +n! 1!x] e−x +n! ∫dx e−x ∫dx xn e−x =−n! e−x ∑k=0 n xkk!$

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

$∫dx xn e−x =−Γ( n+1,x)$

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

$∫−∞ ∞dx e−ax2 =πa$

Square the original integral and change to polar coordinates:

$∫−∞ ∞dx e−ax2 =[∫−∞ ∞dx e−ax2 ×∫−∞ ∞dy e−ay2 ]12 ∫−∞ ∞dx e−a x2 =[∫0 ∞rdr ∫02π dφ e−ar2 ]12 ∫−∞ ∞dx e−a x2 =[π ∫0 ∞dr2 e−ar2 ]12 ∫−∞ ∞dx e−ax2 =πa$

$∫−∞ ∞dx eiax2 =iπa$

Substitute $-ia$ for $a$ in the previous result.

$∫−∞ ∞dx e−ax2 +bx= πa exp(b2 4a)$

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

$∫−∞ ∞dx e−ax2 +bx =∫−∞ ∞dx exp[−a(x -b2a )2 +b2 4a] ∫−∞ ∞dx e−ax2 +bx =exp(b2 4a) ∫−∞ ∞dy e−a y2 ∫−∞ ∞dx e−ax2 +bx= πa exp(b2 4a)$

# 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