nyuuzyou/wikis huggingface dataset:
Definite Integral to Infinity of Cosine of a x over Hyperbolic Cosine of b x

== Theorem ==
:$\ds \int_0^\infty \frac {\cos a x} {\cosh b x} \rd x = \frac \pi {2 b} \sech \frac {a \pi} {2 b}$
where:
:$a$ and $b$ are positive real numbers
:$\sech$ denotes the hyperbolic secant function.
== Proof ==
 
We have, as $b, n > 0$:
We similarly have:
So:
By Mittag-Leffler Expansion for Hyperbolic Secant Function:
:$\ds \pi \map \sech {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 z^2}$
Setting $z = \dfrac a {2 b}$ gives: 
:$\ds \pi \map \sech {\frac {a \pi} {2 b} } = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 \paren {\frac a {2 b} }^2}$
We therefore have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.113$
Category:Definite Integrals involving Cosine Function
Category:Definite Integrals involving Hyperbolic Cosine Function
Definite Integral to Infinity of Cosine p x minus Cosine q x over x Squared

== Theorem ==
:$\ds \int_0^\infty \frac {\cos p x - \cos q x} {x^2} \rd x = \frac {\pi \paren {\size q - \size p} } 2$
where $p, q$ are real numbers.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.39$
Category:Definite Integrals involving Cosine Function
Definite Integral to Infinity of Cosine x over Root x

== Theorem ==
:$\ds \int_0^\infty \frac {\cos x} {\sqrt x} \rd x = \sqrt {\frac \pi 2}$
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.53$
Category:Definite Integrals involving Cosine Function
Definite Integral to Infinity of Cube of Sine x over x Cubed

== Theorem ==
:$\ds \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \frac {3 \pi} 8$
== Proof ==
Let: 
:$\ds \map I \alpha = \int_0^\infty \frac {\map {\sin^3} {\alpha x} } {x^3} \rd x$
for positive real parameter $\alpha$.
We have: 
We aim to evaluate explicitly: 
:$\ds \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \map I 1$
Differentiating with respect to $\alpha$ we have: 
We therefore have: 
By Fundamental Theorem of Calculus: Second Part, we also have: 
:$\ds \int_0^1 \map {I'} \alpha \rd \alpha = \map I 1 - \map I 0 = \map I 1$
giving: 
:$\ds \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \frac {3 \pi} 8$ 
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.58$
Category:Definite Integrals involving Sine Function
Definite Integral to Infinity of Exponential of -(a x^2 plus b over x^2)

== Theorem ==
:$\ds \int_0^\infty \map \exp {-\paren {a x^2 + \frac b {x^2} } } \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-2 \sqrt {a b} }$
where $a$ and $b$ are strictly positive real numbers. 
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.78$
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -(a x^2 plus b x plus c)

== Theorem ==
:$\ds \int_0^\infty \map \exp {-\paren {a x^2 + b x + c} } \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {\frac {b^2 - 4 a c} {4 a} } \map \erfc {\frac b {2 \sqrt a} }$
where:
:$\erfc$ denotes the complementary error function
:$a$, $b$ and $c$ are real numbers with $a > 0$.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.74$
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -a x^2

== Theorem ==
:$\ds \int_0^\infty e^{-a x^2} \rd x = \frac 1 2 \sqrt {\frac \pi a}$
where $a$ is a strictly positive real number.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.72$
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -a x^2 by Cosine of b x

== Theorem ==
:$\ds \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$
where $a$ is a strictly positive real number.
== Proof ==
Fix $a$ and define: 
:$\ds \map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x$
for all $b \in \R$. 
Then, we have: 
Note that: 
So:
We then have: 
:$\dfrac {\map {I'} b} {\map I b} = -\dfrac b {2 a}$
Integrating, by Primitive of Function under its Derivative and Primitive of Constant:
:$\ln \size {\map I b} = -\dfrac {b^2} {4 a} + C$
for some $C \in \R$.
So: 
:$\map I b = A \map \exp {-\dfrac {b^2} {4 a} }$
for some $A \in \R$.
We have: 
on the other hand we have: 
So we have: 
:$\ds\map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$
for all $b \in \R$ as required.
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.73$
Category:Definite Integrals involving Exponential Function
Category:Definite Integrals involving Cosine Function
Definite Integral to Infinity of Exponential of -a x by Cosine of b x

== Theorem ==
:$\ds \int_0^\infty e^{-a x} \cos b x \rd x = \frac a {a^2 + b^2}$
 
where $a$ and $b$ are real numbers with $a > 0$. 
== Proof ==
Note that we have, by Linear Combination of Sine and Cosine: 
:$\ds 0 \le \size {\frac {e^{-a x} \paren {-a \cos b x + b \sin x} } {a^2 + b^2} } \le \frac {e^{-a x} \sqrt {a^2 + b^2} } {a^2 + b^2} = \frac {e^{-a x} } {\sqrt {a^2 + b^2} }$
By Exponential Tends to Zero and Infinity: 
:$\ds \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} } {\sqrt {a^2 + b^2} } } = 0$
So by the Squeeze Theorem: 
:$\ds \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {-a \cos b x + b \sin x} } {a^2 + b^2} } = 0$
So: 
:$\ds \int_0^\infty e^{-a x} \cos b x \rd x = \frac a {a^2 + b^2}$
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.68$
Category:Definite Integrals involving Cosine Function
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -a x by One minus Cosine x over x Squared

== Theorem ==
:$\ds \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } {x^2} \rd x = \arccot a - \frac a 2 \map \ln {a^2 + 1} + a \ln a$
where $a$ is a positive real number.
== Proof ==
Set: 
:$\ds \map I a = \int_0^\infty \frac {e^{-a x} \paren {1 - \cos x} } {x^2} \rd x$
for $a > 0$.
We have: 
We then have, from Primitive of $\dfrac 1 x$ and Primitive of $\dfrac x {x^2 + a^2}$:
:$\ds \map {I'} a = \ln a - \frac 1 2 \map \ln {a^2 + 1} + C$
for some constant $C \in \R$. 
Note that: 
On the other hand: 
so:
:$\ds \map {I'} a = \ln a - \frac 1 2 \map \ln {a^2 + 1}$
We have: 
for some constant $C_2 \in \R$.
Similarly we have: 
On the other hand:
We hence have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.89$
Category:Definite Integrals involving Exponential Function
Category:Definite Integrals involving Cosine Function
Definite Integral to Infinity of Exponential of -a x by Sine of b x

== Theorem ==
:$\ds \int_0^\infty e^{-a x} \sin b x \rd x = \frac b {a^2 + b^2}$
 
where $a$ and $b$ are real numbers with $a > 0$. 
== Proof ==
Note that we have, by Linear Combination of Sine and Cosine: 
:$\ds 0 \le \size {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } \le \frac {e^{-a x} \sqrt {a^2 + b^2} } {a^2 + b^2} = \frac {e^{-a x} } {\sqrt {a^2 + b^2} }$
By Exponential Tends to Zero and Infinity: 
:$\ds \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} } {\sqrt {a^2 + b^2} } } = 0$
So by the Squeeze Theorem: 
:$\ds \lim_{x \mathop \to \infty} \paren {\frac {e^{-a x} \paren {a \sin b x + b \cos b x} } {a^2 + b^2} } = 0$
So: 
:$\ds \int_0^\infty e^{-a x} \sin b x \rd x = \frac b {a^2 + b^2}$
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.69$
Category:Definite Integrals involving Sine Function
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -a x by Sine of b x over x

== Theorem ==
:$\ds \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x = \map \arctan {\frac b a}$
where $a$ and $b$ are real number with $a > 0$.
== Proof ==
Take $a$ constant and define: 
:$\ds \map I b = \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x$
We have: 
so:
for some constant $C \in \R$.
We have: 
on the other hand:
so:
:$C = 0$
So we have: 
:$\ds \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x = \map \arctan {\frac b a}$
as required.
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.70$
Category:Definite Integrals involving Sine Function
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x

== Theorem ==
:$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } x \rd x = \ln \frac b a$
 
where $a$ and $b$ are strictly positive real numbers.
== Proof ==
Note that the integrand is of the form: 
:$\ds \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x$
where:
:$\map f x = e^{-x}$
We have, by Derivative of Exponential Function:
:$\map {f'} x = -e^{-x}$
which is continuous on $\R$.
We also have, by Exponential Tends to Zero and Infinity: 
:$\ds \lim_{x \mathop \to \infty} \map f x = \lim_{x \mathop \to \infty} e^{-x} = 0$
As $f$ is continuously differentiable, and $\ds \lim_{x \mathop \to \infty} \map f x$ exists and is finite, we may apply Frullani's Integral, giving: 
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.71$
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x by Cosecant of p x

== Theorem ==
:$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \arctan \frac b p - \arctan \frac a p$
where:
:$a$ and $b$ are non-negative real numbers
:$p$ is a non-zero real number.
== Proof ==
Fix $p$ and set: 
:$\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x$
for all $\alpha \ge 0$.
Then: 
:$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \map I a - \map I b$
We have: 
so:
for all $\alpha \ge 0$ and constant $C \in \R$.
We therefore have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.88$
Category:Definite Integrals involving Exponential Function
Category:Definite Integrals involving Sine Function
Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x by Secant of p x

== Theorem ==
:$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x = \frac 1 2 \map \ln {\frac {b^2 + p^2} {a^2 + p^2} }$
where:
:$a$ and $b$ are non-negative real numbers
:$p$ is a real number.
== Proof ==
Fix $p$ and set: 
:$\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \sec p x} \rd x$
for all $\alpha \ge 0$.
Then: 
:$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x = \map I a - \map I b$
We have: 
so:
for all $\alpha \ge 0$, for some constant $C \in \R$.
We then have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.87$
Category:Definite Integrals involving Exponential Function
Category:Definite Integrals involving Cosine Function
Definite Integral to Infinity of Exponential of -i x^2

== Theorem ==
:$\ds \int_0^\infty \map \exp {-i x^2} \rd x = \frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i}$
== Proof ==
Let $R$ be a positive real number. 
Let $C_1$ be the straight line segment from $0$ to $R$. 
Let $C_2$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $R e^{i \pi/4}$ anticlockwise.
Let $C_3$ be the straight line segment from $R e^{i \pi/4}$ to $0$.
Let $\Gamma = C_1 \cup C_2 \cup C_3$. 
Let: 
:$\map f z = \map \exp {-z^2}$
From Complex Exponential Function is Entire, $f$ is holomorphic along $\Gamma$ and inside the region that it bounds. 
So, by the Cauchy-Goursat Theorem: 
:$\ds \int_\Gamma \map \exp {-z^2} \rd z = 0$
From Contour Integral of Concatenation of Contours, we therefore have: 
We have: 
So, taking $R \to \infty$, we have:
:$\ds e^{i \pi/4} \int_0^\infty \map \exp {-i t^2} \rd t = \int_0^\infty \map \exp {-x^2} \rd x$
giving: 
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Exponential of -x^2 by Logarithm of x

== Theorem ==
:$\ds \int_0^\infty e^{-x^2} \ln x \rd x = -\frac {\sqrt \pi} 4 \paren {\gamma + 2 \ln 2}$
where $\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
Consider the integral: 
:$\ds \int_0^\infty x^t e^{-x^2} \rd x$
for positive real parameter $t$.
Using Definite Integral to Infinity of $x^m e^{-a x^2}$, we have: 
:$\ds \int_0^\infty x^t e^{-x^2} \rd x = \frac 1 2 \map \Gamma {\frac {1 + t} 2}$
Differentiating the $t$ we have: 
Differentiating the using the Chain Rule for Derivatives gives: 
:$\ds \int_0^\infty x^t e^{-x^2} \ln x \rd x = \frac 1 4 \map {\Gamma'} {\frac {1 + t} 2}$
We therefore have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.100$
Category:Definite Integrals involving Exponential Function
Category:Definite Integrals involving Logarithm Function
Category:Euler-Mascheroni Constant
Definite Integral to Infinity of Exponential of -x by Logarithm of x

== Theorem ==
Let $\ln t$ denote the natural logarithm function for real $t > 0$.
Let $e^{-t}$ denote the real exponential.
Then:
:$\ds \int_{0^+}^{\mathop \to +\infty} \ln t \, e^{-t} \rd t = - \gamma$
where the is an improper integral, and $\gamma$ is the Euler-Mascheroni Constant.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.99$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Exponential Function
Category:Gamma Function
Definite Integral to Infinity of Hyperbolic Sine of a x over Exponential of b x minus One

== Theorem ==
:$\ds \int_0^\infty \frac {\sinh a x} {e^{b x} - 1} \rd x = \frac 1 {2 a} - \frac \pi {2 b} \cot \frac {a \pi} b$
where:
:$a$ and $b$ are positive real numbers with $b > a$
:$\cot$ denotes the cotangent function.
== Proof ==
Note that as $b > a$, we have that $a - b < 0$.
As $b > 0$, we therefore have $a - \paren {n + 1} b < 0$ for all positive integer $n$.
We also have that as $a + \paren {n + 1} b > 0$, that $-\paren {a + \paren {n + 1} b} < 0$.
So, by Exponential Tends to Zero and Infinity: 
:$\ds \lim_{x \mathop \to \infty} \frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} = 0$
and: 
:$\ds \lim_{x \mathop \to \infty} \frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} = 0$
We therefore have: 
By Mittag-Leffler Expansion for Cotangent Function, we have: 
:$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$
where $z$ is not an integer. 
Setting $z = \dfrac a b$, we have: 
:$\ds \pi \cot \frac {\pi a} b = \frac b a + 2 \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$
Rearrangement gives:
:$\ds \frac \pi 2 \cot \frac {\pi a} b - \frac b {2 a} = \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$
So: 
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.117$
Category:Definite Integrals involving Hyperbolic Sine Function
Definite Integral to Infinity of Hyperbolic Sine of a x over Exponential of b x plus One

== Theorem ==
:$\ds \int_0^\infty \frac {\sinh a x} {e^{b x} + 1} \rd x = \frac \pi {2 b} \csc \frac {a \pi} b - \frac 1 {2 a}$
where:
:$a$ and $b$ are positive real numbers with $b > a$
:$\csc$ is the cosecant function.
== Proof ==
Note that as $b > a$, we have that $a - b < 0$.
As $b > 0$, we therefore have $a - \paren {n + 1} b < 0$ for all positive integer $n$.
We also have that as $a + \paren {n + 1} b > 0$, that $-\paren {a + \paren {n + 1} b} < 0$.
So, by Exponential Tends to Zero and Infinity: 
:$\ds \lim_{x \mathop \to \infty} \frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} = 0$
and: 
:$\ds \lim_{x \mathop \to \infty} \frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} = 0$
We therefore have:
We have by Mittag-Leffler Expansion for Cosecant Function:
:$\ds \pi \map \csc {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 - n^2}$ 
Setting $z = \dfrac a b$ we have: 
:$\ds \pi \map \csc {\frac {\pi a} b} = \frac b a + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$
Rearranging gives: 
:$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} = \frac \pi 2 \map \csc {\frac {\pi a} b} - \frac b {2 a}$
Therefore: 
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.116$
Category:Definite Integrals involving Hyperbolic Sine Function
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Logarithm of Exponential of x plus One over Exponential of x minus One

== Theorem ==
:$\ds \int_0^\infty \map \ln {\frac {e^x + 1} {e^x - 1} } \rd x = \frac {\pi^2} 4$
== Proof ==
We can write: 
Let: 
:$u = \coth \dfrac x 2$
We have, by Derivative of Hyperbolic Cotangent Function:
:$\dfrac {\d u} {\d x} = -\dfrac 1 2 \csch^2 \dfrac x 2$
From Difference of Squares of Hyperbolic Cotangent and Cosecant, this can be written: 
:$\dfrac {\d u} {\d x} = \dfrac 1 2 \paren {1 - \coth^2 \dfrac x 2} = \dfrac 1 2 \paren {1 - u^2}$
From Limit to Infinity of Hyperbolic Cotangent Function, we have: 
:as $x \to \infty$, $u \to 1$.
We also have: 
:as $x \to 0^+$, $u \to \infty$.
With this, we have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.101$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Power of x by Exponential of -a x^2

== Theorem ==
:$\ds \int_0^\infty x^m e^{-a x^2} \rd x = \frac {\map \Gamma {\paren {m + 1}/2} } {2 a^{\paren {m + 1}/2} }$
where $m$ and $a$ are real numbers with $m > -1$ and $a > 0$.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.77$
Category:Definite Integrals involving Exponential Function
Definite Integral to Infinity of Power of x by Logarithm of x over One plus x

== Theorem ==
:$\ds \int_0^\infty \frac {x^{p - 1} \ln x} {1 + x} \rd x = -\pi^2 \csc p \pi \cot p \pi$
where: 
:$p$ is a real number with $0 < p < 1$
:$\csc$ denotes the cosecant function
:$\cot$ denotes the cotangent function.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.97$
Category:Definite Integrals involving Logarithm Function
Definite Integral to Infinity of Power of x over 1 + 2 x Cosine Beta + x Squared

== Theorem ==
:$\ds \int_0^\infty \dfrac {x^m \rd x} {1 + 2 x \cos \beta + x^2} = \frac \pi {\sin m x} \frac {\sin m \beta} {\sin \beta}$
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Rational or Irrational expressions: $15.21$
Category:Examples of Definite Integrals
Definite Integral to Infinity of Power of x over 1 + x

== Theorem ==
:$\ds \int_0^\infty \dfrac {x^{p - 1} \rd x} {1 + x} = \frac \pi {\sin \pi p}$
for $0 < p < 1$. 
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 15$: Definite Integrals involving Rational or Irrational expressions: $15.19$
Category:Definite Integral to Infinity of Power of x over 1 + x
Category:Examples of Definite Integrals
Definite Integral to Infinity of Power of x over Hyperbolic Sine of a x

== Theorem ==
:$\ds \int_0^\infty \frac {x^n} {\sinh a x} \rd x = \frac {2^{n + 1} - 1} {2^n a^{n + 1} } \map \Gamma {n + 1} \map \zeta {n + 1}$
where:
:$a$ and $n$ are positive real numbers
:$\Gamma$ denotes the gamma function
:$\zeta$ denotes the Riemann zeta function.
== Proof ==
We have: 
So: 
giving: 
:$\ds \int_0^\infty \frac {x^n} {\sinh a x} \rd x = \frac {2^{n + 1} - 1} {2^n a^{n + 1} } \map \Gamma {n + 1} \map \zeta {n + 1}$
== Sources ==
* : $\S 15$: Definite Integrals involving Hyperbolic Functions: $15.115$
Category:Definite Integrals involving Hyperbolic Sine Function
Category:Gamma Function
Category:Riemann Zeta Function
Definite Integral to Infinity of Power of x over Power of x plus Power of a

== Theorem ==
:$\ds \int_0^\infty \dfrac {x^m \rd x} {x^n + a^n} = \frac {\pi a^{m + 1 - n} } {n \map \sin {\paren {m + 1} \frac \pi n} }$
for $0 < m + 1 < n$.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Rational or Irrational expressions: $15.20$
Category:Examples of Definite Integrals
Definite Integral to Infinity of Reciprocal of 1 plus Power of x

== Theorem == 
:$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$
where:
:$n$ is a real number greater than 1
:$\csc$ is the cosecant function.
=== Corollary ===
== Proof 1 ==
 
== Proof 2 ==
 
== Proof 3 ==
 
== Proof 4 ==
 
Category:Definite Integral to Infinity of Reciprocal of 1 plus Power of x
Category:Examples of Definite Integrals
Definite Integral to Infinity of Reciprocal of 1 plus Power of x/Corollary

== Theorem == 
:$\ds \int_0^\infty \frac 1 {a^n + x^n} \rd x = \frac \pi {n a^{n - 1} } \map \csc {\frac \pi n}$
where:
:$n$ is a real number greater than 1
:$\csc$ is the cosecant function
:$a \ne 0$.
== Proof ==
Category:Definite Integral to Infinity of Reciprocal of 1 plus Power of x
Definite Integral to Infinity of Reciprocal of 1 plus Power of x/Proof 1

== Theorem == 
== Proof ==
From Euler's Reflection Formula:
:$\map \Gamma {\dfrac 1 n} \map \Gamma {1 - \dfrac 1 n} = \pi \map \csc {\dfrac \pi n}$
Then: 
Note we have: 
As $\theta \to \dfrac \pi 2$, $\tan \theta \to \infty$ and $\tan 0 = 0$, so making a substitution of $x = \paren {\tan \theta}^{\frac 2 n}$ to our original integral:
So we have: 
:$\ds \pi \map \csc {\frac \pi n} = n \int_0^\infty \frac 1 {1 + x^n} \rd x$
Hence:
:$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$
Category:Definite Integral to Infinity of Reciprocal of 1 plus Power of x
Definite Integral to Infinity of Reciprocal of 1 plus Power of x/Proof 2

== Theorem == 
== Proof == 
Let $R > 1$ be a real number.
Let:
:$\ds C_R = \set {R e^{i \theta} : 0 \le \theta \le \frac {2 \pi} n}$
Let $L_R$ be the straight line segment from $0$ to $R e^{\frac {2 \pi i} n}$.
Let $\Gamma_R = \closedint 0 R \cup C_R \cup L_R$, traversed anticlockwise. 
Then: 
We can show the second integral to vanish as $R \to \infty$: 
So we have: 
:$\ds \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z = \paren {1 - e^{\frac {2 \pi i} n} } \int_0^\infty \frac 1 {1 + x^n} \rd x$
We have that the integrand is meromorphic with a single simple pole at $z = e^{\frac \pi n i}$.
This is contained within the contour
Hence Cauchy's Residue Theorem can be applied to evaluate the :
So: 
:$\ds \paren {e^{\frac {2 \pi i} n} - 1} \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac {2 \pi i e^{i \frac \pi n} } n$
giving: 
Category:Definite Integral to Infinity of Reciprocal of 1 plus Power of x
Definite Integral to Infinity of Reciprocal of Exponential of x minus One minus Exponential of -x over x

== Theorem ==
:$\ds \int_0^\infty \paren {\frac 1 {e^x - 1} - \frac {e^{-x} } x} \rd x = \gamma$
where $\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
Let: 
:$x = e^{-t}$
By Derivative of $e^{a x}$:
:$\dfrac {\d x} {\d t} = -e^{-t}$
By Exponential of Zero, we have: 
:as $x \to 1$, $t \to 0$.
By Exponential Tends to Zero and Infinity, we have: 
:as $x \to 0$, $t \to \infty$.
We therefore have: 
== Also see ==
* Gauss's Integral Form of Digamma Function
== Sources ==
* : $\S 15$: Definite Integrals involving Exponential Functions: $15.86$
Category:Definite Integrals involving Exponential Function
Category:Euler-Mascheroni Constant
Definite Integral to Infinity of Reciprocal of x Squared plus a Squared

== Theorem ==
:$\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
for $a \ne 0$. 
=== Corollary ===
== Proof 1 ==
 
== Proof 2 ==
 
== Proof 3 ==
 
== Sources ==
* : $\S 15$: Definite Integrals Involving Rational or Irrational Expressions: $15.18$
* : $\S 18$: Definite Integrals: Definite Integrals Involving Rational or Irrational Expressions: $18.18.$
Category:Definite Integral to Infinity of Reciprocal of x Squared plus a Squared
Category:Definite Integrals involving x squared plus a squared