A Miscellany of Useful Integrals

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

$\int d x (1 - a x^{m})^{r} = x {}_{2}F_{1} (- r, \frac{1}{m}; \frac{1}{m} + 1; a x^{m})$

Given the general binomial expansion

$\begin{array}{l} (1 - x)^{r} = 1 - r x + \frac{r (r - 1)}{2!} x^{2} - \frac{r (r - 1) (r - 2)}{3!} x^{3} + \dots \\ = \sum_{k = 0}^{\infty} \frac{Γ (- r + k)}{Γ (- r)} \frac{x^{k}}{k!} \end{array}$

the evaluation of the integral is

$\begin{array}{l} \int d x (1 - a x^{m})^{r} = \int d x \sum_{k = 0}^{\infty} \frac{Γ (- r + k)}{Γ (- r)} \frac{a^{k}}{k!} x^{m k} \\ = \sum_{k = 0}^{\infty} \frac{Γ (- r + k)}{Γ (- r)} \frac{a^{k}}{k!} \frac{x^{m k + 1}}{m k + 1} \\ = \sum_{k = 0}^{\infty} \frac{Γ (- r + k)}{Γ (- r)} \frac{\frac{1}{m}}{\frac{1}{m} + k} \frac{a^{k}}{k!} x^{m k + 1} \\ = \sum_{k = 0}^{\infty} \frac{Γ (- r + k)}{Γ (- r)} \frac{Γ (\frac{1}{m} + k)}{Γ (\frac{1}{m})} \frac{Γ (\frac{1}{m} + 1)}{Γ (\frac{1}{m} + 1 + k)} \frac{a^{k}}{k!} x^{m k + 1} \\ \int d x (1 - a x^{m})^{r} = x {}_{2}F_{1} (- r, \frac{1}{m}; \frac{1}{m} + 1; a x^{m}) \end{array}$

using $Γ (x + 1) = x Γ (x)$ 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.

$\int d x x^{p} (1 - a x^{m})^{r} = \frac{x^{p + 1}}{p + 1} {}_{2}F_{1} (- r, \frac{p + 1}{m}; \frac{p + 1}{m} + 1; a x^{m})$

Change the variable of integration and use the previous result:

$\begin{array}{l} \int d x x^{p} (1 - a x^{m})^{r} = \frac{1}{p + 1} \int d (x^{p + 1}) [1 - a (x^{p + 1})^{m / (p + 1)}]^{r} \\ = \frac{x^{p + 1}}{p + 1} {}_{2}F_{1} (- r, \frac{p + 1}{m}; \frac{p + 1}{m} + 1; a x^{m}) \end{array}$

$\int d x (x^{p} - a x^{m})^{r} = \frac{x^{p r + 1}}{p r + 1} {}_{2}F_{1} (- r, \frac{p r + 1}{m - p}; \frac{p r + 1}{m - p} + 1; a x^{m - p})$

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

$\begin{array}{l} \int d x (x^{p} - a x^{m})^{r} = \int d x x^{p r} (1 - a x^{m - p})^{r} \\ = \frac{x^{p r + 1}}{p r + 1} {}_{2}F_{1} (- r, \frac{p r + 1}{m - p}; \frac{p r + 1}{m - p} + 1; a x^{m - p}) \end{array}$

$\begin{array}{l} \int d x x^{p} (1 - a_{1} x^{m})^{r_{1}} \dots (1 - a_{n} x^{m})^{r_{n}} \\ = \frac{x^{p + 1}}{p + 1} F_{D}^{(n)} (\frac{p + 1}{m}; - r_{1}, \dots, - r_{n}; \frac{p + 1}{m} + 1; a_{1} x^{m}, \dots, a_{n} x^{m}) \end{array}$

Given the general binomial expansion

the evaluation of the integral is

$\begin{array}{l} \int d x x^{p} (1 - a_{1} x^{m})^{r_{1}} \dots (1 - a_{n} x^{m})^{r_{n}} \\ = \int d x \sum_{k_{1}, \dots, k_{n} = 0}^{\infty} \frac{Γ (- r_{1} + k_{1})}{Γ (- r_{1})} \dots \frac{Γ (- r_{n} + k_{n})}{Γ (- r_{n})} \frac{a_{1}^{k_{1}}}{k_{1}!} \dots \frac{a_{n}^{k_{n}}}{k_{n}!} x^{m k_{1} + \dots + m k_{n} + p} \\ = x^{p + 1} \sum_{k_{1}, \dots, k_{n} = 0}^{\infty} \frac{Γ (- r_{1} + k_{1})}{Γ (- r_{1})} \dots \frac{Γ (- r_{n} + k_{n})}{Γ (- r_{n})} \frac{\frac{1}{m}}{k_{1} + \dots + k_{n} + \frac{p + 1}{m}} \\ \times \frac{(a_{1} x^{m})^{k_{1}}}{k_{1}!} \dots \frac{(a_{n} x^{m})^{k_{n}}}{k_{n}!} \\ = \frac{x^{p + 1}}{p + 1} \sum_{k_{1}, \dots, k_{n} = 0}^{\infty} \frac{Γ (- r_{1} + k_{1})}{Γ (- r_{1})} \dots \frac{Γ (- r_{n} + k_{n})}{Γ (- r_{n})} \\ \times \frac{Γ (\frac{p + 1}{m} + k_{1} + \dots + k_{n})}{Γ (\frac{p + 1}{m})} \frac{Γ (\frac{p + 1}{m} + 1)}{Γ (\frac{p + 1}{m} + 1 + k_{1} + \dots + k_{n})} \\ \times \frac{(a_{1} x^{m})^{k_{1}}}{k_{1}!} \dots \frac{(a_{n} x^{m})^{k_{n}}}{k_{n}!} \\ = \frac{x^{p + 1}}{p + 1} F_{D}^{(n)} (\frac{p + 1}{m}; - r_{1}, \dots, - r_{n}; \frac{p + 1}{m} + 1; a_{1} x^{m}, \dots, a_{n} x^{m}) \end{array}$

using $Γ (x + 1) = x Γ (x)$ 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.

$\begin{array}{l} \int d x x^{p} (1 - x)^{q} (1 - a_{1} x)^{r_{1}} \dots (1 - a_{n} x)^{r_{n}} \\ = B (p + 1, q + 1) x^{p + 1} F_{D}^{(n)} (p + 1; - r_{1}, \dots, - r_{n}; p + q + 2; a_{1} x, \dots, a_{n} x) \end{array}$

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

$\begin{array}{l} \int d x x^{p} (1 - x)^{q} (1 - a_{1} x)^{r_{1}} \dots (1 - a_{n} x)^{r_{n}} \\ = \int d x \sum_{k, k_{1}, \dots, k_{n} = 0}^{\infty} \frac{Γ (- r_{1} + k_{1})}{Γ (- r_{1})} \dots \frac{Γ (- r_{n} + k_{n})}{Γ (- r_{n})} \frac{a_{1}^{k_{1}}}{k_{1}!} \dots \frac{a_{n}^{k_{n}}}{k_{n}!} \\ \times \frac{1}{k!} \frac{Γ (- q + k)}{Γ (- q)} x^{k + k_{1} + \dots + k_{n} + p} \\ = x^{p + 1} \sum_{k, k_{1}, \dots, k_{n} = 0}^{\infty} \frac{Γ (- r_{1} + k_{1})}{Γ (- r_{1})} \dots \frac{Γ (- r_{n} + k_{n})}{Γ (- r_{n})} \frac{(a_{1} x)^{k_{1}}}{k_{1}!} \dots \frac{(a_{n} x)^{k_{n}}}{k_{n}!} \\ \times \frac{1}{k!} \frac{Γ (- q + k)}{Γ (- q)} \frac{1}{k + k_{1} + \dots + k_{n} + p + 1} \end{array}$

Complete the summation over k

$\begin{array}{l} \sum_{k = 0}^{\infty} \frac{1}{k!} \frac{Γ (- q + k)}{Γ (- q)} \frac{1}{k + k_{1} + \dots + k_{n} + p + 1} \\ = \sum_{k = 0}^{\infty} \frac{1}{k!} \frac{Γ (- q + k)}{Γ (- q)} \frac{Γ (k + k_{1} + \dots + k_{n} + p + 1)}{Γ (k + k_{1} + \dots + k_{n} + p + 2)} \\ = \frac{Γ (k_{1} + \dots + k_{n} + p + 1)}{Γ (k_{1} + \dots + k_{n} + p + 2)} \\ \times {}_{2}F_{1} (- q, k_{1} + \dots + k_{n} + p + 1; k_{1} + \dots + k_{n} + p + 2; 1) \\ = \frac{Γ (k_{1} + \dots + k_{n} + p + 1) Γ (q + 1)}{Γ (k_{1} + \dots + k_{n} + p + q + 2)) Γ (1)} \end{array}$

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

${}_{2}F_{1} (a, b; c; 1) = \frac{Γ (c) Γ (c - a - b)}{Γ (c - a) Γ (c - b)}$

in the third. The final integral is

$\begin{array}{l} \int d x x^{p} (1 - x)^{q} (1 - a_{1} x)^{r_{1}} \dots (1 - a_{n} x)^{r_{n}} \\ = \frac{Γ (p + 1) Γ (q + 1)}{Γ (p + q + 2)} x^{p + 1} F_{D}^{(n)} (p + 1; - r_{1}, \dots, - r_{n}; p + q + 2; a_{1} x, \dots, a_{n} x) \\ = B (p + 1, q + 1) x^{p + 1} F_{D}^{(n)} (p + 1; - r_{1}, \dots, - r_{n}; p + q + 2; a_{1} x, \dots, a_{n} x) \end{array}$

For $q = 0$ this result is identical to setting $m = 1$ in the previous result.

$\begin{array}{l} \int_{0}^{1} d x x^{a - 1} (1 - x)^{c - a - 1} (1 - z_{1} x)^{- b_{1}} (1 - z_{2} x)^{- b_{2}} (1 - z_{3} x)^{- b_{3}} \\ = B (a, c - a) F_{D} (a; b_{1}, b_{2}, b_{3}; c; z_{1}, z_{2}, z_{3}) \end{array}$

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

$\begin{array}{l} \int_{0}^{1} d x x^{a - 1} (1 - x)^{c - a - 1} (1 - z_{1} x)^{- b_{1}} (1 - z_{2} x)^{- b_{2}} \\ = B (a, c - a) F_{1} (a; b_{1}, b_{2}; c; z_{1}, z_{2}) \end{array}$

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

$\int_{0}^{1} d x x^{a - 1} (1 - x)^{c - a - 1} (1 - z x)^{- b} = B (a, c - a) {}_{2}F_{1} (a, b; c; z)$

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

$\int d θ {sin}^{p} θ {cos}^{m} θ = \frac{{sin}^{p + 1} θ}{p + 1} {}_{2}F_{1} (\frac{1 - m}{2}, \frac{p + 1}{2}; \frac{p + 3}{2}; {sin}^{2} θ)$

Change variable to $x = sin θ$ and use a binomial integral:

$\begin{array}{l} \int d θ {sin}^{p} θ {cos}^{m} θ = \int d x x^{p} (1 - x^{2})^{(m - 1) / 2} \\ = \frac{{sin}^{p + 1} θ}{p + 1} {}_{2}F_{1} (\frac{1 - m}{2}, \frac{p + 1}{2}; \frac{p + 3}{2}; {sin}^{2} θ) \end{array}$

$\int d θ {sin}^{p} θ {cos}^{m} θ = - \frac{{cos}^{m + 1} θ}{m + 1} {}_{2}F_{1} (\frac{1 - p}{2}, \frac{m + 1}{2}; \frac{m + 3}{2}; {cos}^{2} θ)$

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

$\int_{0}^{π / 2} d θ {sin}^{p} θ {cos}^{m} θ = \frac{Γ (\frac{p + 1}{2}) Γ (\frac{m + 1}{2})}{2 Γ (\frac{p + m}{2} + 1)}$

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

$B (x, y) = \frac{Γ (x) Γ (y)}{Γ (x + y)} = 2 \int_{0}^{π / 2} d θ {sin}^{2 x - 1} θ {cos}^{2 y - 1} θ$

the integral is trivial. Alternately given the Gauss summation formula

${}_{2}F_{1} (a, b; c; 1) = \frac{Γ (c) Γ (c - a - b)}{Γ (c - a) Γ (c - b)}$

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

$\begin{array}{l} \int_{0}^{π / 2} d θ {sin}^{p} θ {cos}^{m} θ = \frac{1}{p + 1} {}_{2}F_{1} (\frac{1 - m}{2}, \frac{p + 1}{2}; \frac{p + 3}{2}; 1) \\ = \frac{Γ (\frac{p + 3}{2}) Γ (\frac{m + 1}{2})}{(p + 1) Γ (\frac{p + m}{2} + 1) Γ (1)} = \frac{Γ (\frac{p + 1}{2}) Γ (\frac{m + 1}{2})}{2 Γ (\frac{p + m}{2} + 1)} \end{array}$

$\begin{array}{l} \int_{0}^{π / 2} d θ {sin}^{p} θ {cos}^{m} θ = \frac{1}{m + 1} {}_{2}F_{1} (\frac{1 - p}{2}, \frac{m + 1}{2}; \frac{m + 3}{2}; 1) \\ = \frac{Γ (\frac{m + 3}{2}) Γ (\frac{p + 1}{2})}{(m + 1) Γ (\frac{m + p}{2} + 1) Γ (1)} = \frac{Γ (\frac{p + 1}{2}) Γ (\frac{m + 1}{2})}{2 Γ (\frac{p + m}{2} + 1)} \end{array}$

$\int \frac{d t}{1 + a cos b t} = \frac{2}{b \sqrt{1 - a^{2}}} {tan}^{- 1} [\sqrt{\frac{1 - a}{1 + a}} tan \frac{b t}{2}]$

Given the trigonometric substitution

$\begin{matrix} u = tan (\frac{x}{2}) & sin x = \frac{2 u}{1 + u^{2}} \\ d x = \frac{2 d u}{1 + u^{2}} & cos x = \frac{1 - u^{2}}{1 + u^{2}} \end{matrix}$

set $x = b t$

$\int \frac{d t}{1 + a cos b t} = \frac{2}{b (1 + a)} \int \frac{d u}{1 + \frac{1 - a}{1 + a} u^{2}} = \frac{2}{b \sqrt{1 - a^{2}}} {tan}^{- 1} [\sqrt{\frac{1 - a}{1 + a}} tan \frac{b t}{2}]$

and recognize the integral of an inverse tangent.

Exponential Integrals

$\int d x x^{n} e^{- x} = - n! e^{- x} \sum_{k = 0}^{n} \frac{x^{k}}{k!} \equiv - n! e^{- x} e_{n} (x)$

Integrate by parts n times:

$\begin{array}{l} \int d x x^{n} e^{- x} = - x^{n} e^{- x} + n \int d x x^{n - 1} e^{- x} \\ = - [x^{n} + n x^{n - 1}] e^{- x} + n (n - 1) \int d x x^{n - 2} e^{- x} \\ = - [x^{n} + \frac{n!}{(n - 1)!} x^{n - 1} + \dots + \frac{n!}{1!} x] e^{- x} + n! \int d x e^{- x} \\ \int d x x^{n} e^{- x} = - n! e^{- x} \sum_{k = 0}^{n} \frac{x^{k}}{k!} \end{array}$

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

$\int d x x^{n} e^{- x} = - Γ (n + 1, x)$

since the exponential vanishes at the upper limit.

Substitute $- x$ for $x$ in the previous result.

Substitute $a x$ for $x$ in the previous result.

Gaussian Integrals

Square the original integral and change to polar coordinates:

$\begin{array}{l} \int_{- \infty}^{\infty} d x e^{- a x^{2}} = [\int_{- \infty}^{\infty} d x e^{- a x^{2}} \times \int_{- \infty}^{\infty} d y e^{- a y^{2}}]^{\frac{1}{2}} \\ = [\int_{0}^{\infty} r d r \int_{0}^{2 π} d φ e^{- a r^{2}}]^{\frac{1}{2}} \\ = [π \int_{0}^{\infty} d r^{2} e^{- a r^{2}}]^{\frac{1}{2}} \\ \int_{- \infty}^{\infty} d x e^{- a x^{2}} = \sqrt{\frac{π}{a}} \end{array}$

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

$\int_{- \infty}^{\infty} d x e^{- a x^{2} + b x} = \sqrt{\frac{π}{a}} exp (\frac{b^{2}}{4 a})$

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

$\begin{array}{l} \int_{- \infty}^{\infty} d x e^{- a x^{2} + b x} = \int_{- \infty}^{\infty} d x exp [- a (x - \frac{b}{2 a})^{2} + \frac{b^{2}}{4 a}] \\ = exp (\frac{b^{2}}{4 a}) \int_{- \infty}^{\infty} d y e^{- a y^{2}} \\ \int_{- \infty}^{\infty} d x e^{- a x^{2} + b x} = \sqrt{\frac{π}{a}} exp (\frac{b^{2}}{4 a}) \end{array}$

Miscellaneous Integrals

$\int_{c}^{x} d t \int_{c}^{t} d t \dots \int_{c}^{t} y (t) d t = \frac{1}{(n - 1)!} \int_{c}^{x} (x - t)^{n - 1} y (t) d t$

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

$\frac{d}{d x} \int_{a (x)}^{b (x)} f (x, t) d t = f [x, b (x)] \frac{d b (x)}{d x} - f [x, a (x)] \frac{d a (x)}{d x} + \int_{a (x)}^{b (x)} \frac{\partial f (x, t)}{\partial x} d t$

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:

$\frac{d}{d x} \int_{c}^{x} d t \int_{c}^{t} d t \dots \int_{c}^{t} y (t) d t = \int_{c}^{x} d t \dots \int_{c}^{t} y (t) d t$

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

$\frac{1}{(n - 1)!} \frac{d}{d x} \int_{c}^{x} (x - t)^{n - 1} y (t) d t = \frac{1}{(n - 2)!} \int_{c}^{x} (x - t)^{n - 2} y (t) d t$

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

Contents

Binomial Integrals

Trigonometric Integrals

Exponential Integrals

Gaussian Integrals

Miscellaneous Integrals