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