Problem  According to , the Gamma function is computed as  $$ \Gamma(z)=\int_{0}^{\infty} x^{z-1} e^{-x} d x $$ where $z$ is a complex number.  Using integration by parts, it is easy to verify that $\Gamma(z)=(z-1)\Gamma(z-1)$ and therefore I have $$\Gamma(z) = (z-1)!$$ However, I am not sure how do I compute something like $4.7!$ (or value of any real number). I tried $4.7\cdot 3.7\cdots \cdot 0.7$, but it does not give the correct result as verified by scipy.math.gamma function.
75593
All I know of the Gamma function is $$\Gamma(\alpha) = \int\limits_{0}^{\infty}x^{\alpha - 1}e^{-x}\text{ d}x$$ and the recursive formula $$\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha - 1)\text{.}$$ A friend asked me to try to find $\Gamma(2.7)$. I assume this involves complex analysis, which I have absolutely no background on (except for the basic Euler formula). Using Wolfram, I found that  $$\Gamma(2.7) = e^{0.434821}$$ which suggests to me perhaps writing out $e^{-x}$ as a Taylor series: $$\Gamma(\alpha) = \int\limits_{0}^{\infty}x^{\alpha - 1}\sum\limits_{i=1}^{\infty}\dfrac{(-x)^{i}}{i!}\text{ d}x$$ but alas, this doesn't look helpful.  How does one derive that $\Gamma(2.7) = e^{0.434821}$?
51806
Hi I am trying to prove$$ I:=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x =\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0. $$ I am not sure whether to use an integral representation or to somehow use the Euler reflection formula $$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z} $$ since a previous post used  that to solve  this kind of integral.  Other than this method, we can use the integral representation  $$ \Gamma(t)=\int_0^\infty x^{t-1} e^{-x}\, dx. $$  Also note $\Gamma(n)=(n-1)!$.
5542
I would like to show that  $$ \int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx = \frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}$$  thanks to the beta function which I am not used to handling...  $$\frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}=\frac{2}{5}B(1/2,1/6)=\frac{2}{5} \int_0^{\infty} \frac{ \mathrm dt}{\sqrt{t}(1+t)^{2/3}}$$  ...?
90984
Definitionally, we have that  $$ \int_\gamma f\ \overline{dz} = \overline{\int_\gamma \overline{f}\ dz} $$  Now let $\int_\gamma f\ dz = w = x +yi$.  Question 1: Is it not the case that $\int_\gamma \overline{f}\ dz = \bar{w} = x - iy$?  Question 2:  Assuming that $\int_\gamma \overline{f}\ dz = \bar{w} = x - iy$, then is it not the case that  $$ \overline{\int_\gamma \overline{f}\ dz} = \overline{x - iy} = x+iy = \int_\gamma f\ dz? $$  Basically, I'm failing to see how $\overline{\int_\gamma \overline{f}\ dz}$ isn't just $\int_\gamma f\ dz$.
45248
I got a clock as a gift recently. It has a very novel face in that the hour positions are given by a complex formula. For the most part, I have been able to verify the calculations presented as accurate, but the two o'clock identity has me stumped.  $$ \frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}} $$  I know that $\gamma$ is the . And WolframAlpha tells me that $\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}$ which  makes sense because  $$ \frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}}=\frac{\gamma}{\displaystyle{\frac{\gamma}{2}}}=2 $$  It is not clear how I would show $\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}$. Could anyone shed some light on this or point me to a source (book, article, etc...) where I can read up on this. The usual internet (listed above) resources have not been helpful to me.
22891
In the solution of the so-called , one of the key steps is to compute $$\int_0^1 x^n (\log x)^n~\mathrm dx$$ using the change of variables $x = \exp\left(-\frac{u}{n+1}\right)$ to obtain the Gamma function.  This substitution to me, looks like it was pulled out of thin air. Can someone help me motivate it? How would I have thought of this substitution?
113629
I came across this integral $\int e^{x^n} dx $ and was just wondering how you go about solving it. In this  there approach is to let $y = -x^n$, then $$ dy = -nx^{n-1} dx = -n(-y)^{\frac{n-1}{n}} \ dx \ \Rightarrow \ dx = -\frac{1}{n}(-y)^{\frac{1-n}{n}} \ dy, $$ then sub into the integral, we have $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n}\int_{-\infty}^\infty  y^{\frac{1}{n}-1} e^{-y} \ dy, $$ then they referenced the primary definition of incomplete gamma function, $$ \Gamma(a,y) = \int_y^\infty t^{a-1} e^{-t} \ dt, $$ and claim that we can write our integral in terms of $\Gamma$ function, and the end result is $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, y). $$ Can anyone fill in the missing step for me? Our integral does look very similar to the gamma function, but what happened to the lower bound of the integral? In our integral the lower bound is $-\infty$; but to equate it to the gamma function we need the lower bound to be $y$.   Anyway substitute back the definition of $y$ we have the final result $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, -x^n). $$ I̶'̶m̶ ̶g̶u̶e̶s̶s̶i̶n̶g̶ ̶t̶h̶e̶i̶r̶ ̶a̶n̶s̶w̶e̶r̶ ̶i̶s̶ ̶c̶o̶r̶r̶e̶c̶t̶ ̶b̶e̶c̶a̶u̶s̶e̶ ̶M̶a̶t̶h̶e̶m̶a̶t̶i̶c̶a̶ ̶a̶l̶s̶o̶ ̶g̶i̶v̶e̶s̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶r̶e̶s̶u̶l̶t̶.̶ ̶B̶u̶t̶ ̶a̶n̶o̶t̶h̶e̶r̶ ̶t̶h̶i̶n̶g̶ ̶I̶ ̶d̶o̶n̶'̶t̶ ̶u̶n̶d̶e̶r̶s̶t̶a̶n̶d̶ ̶i̶s̶ ̶w̶h̶y̶ ̶d̶o̶e̶s̶ ̶t̶h̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶l̶ ̶d̶e̶p̶e̶n̶d̶s̶ ̶o̶n̶ ̶$̶x̶$̶?̶ ̶Y̶o̶u̶'̶d̶ ̶t̶h̶i̶n̶k̶ ̶$̶x̶$̶ ̶s̶h̶o̶u̶l̶d̶ ̶b̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶t̶e̶d̶ ̶o̶v̶e̶r̶.̶ Any help is appreciated, thanks.
114152
I'm trying to calculate the following integral:   $$ \int_0^1 e^{-\lambda(1-x)} (1-x)^{n-1} x^{k-n} dx $$ It seems like kind of a combination of Gamma and Beta function. I'm suspecting that it has something to do with both of them, however I can't see the connection.  Any hint would be welcome!
116747
If this is a duplicate in any way, I'm very sorry.  I'm brushing up on some Complex Analysis with Special Functions in mind. Here's a problem I'm stuck on.     Evaluate the integral $$I=\frac{1}{2\pi i}\oint_{\lvert z\rvert=1}f(z)\, dz$$ over the counterclockwise oriented unit circle centred at $0$, where $$f(z)=\frac{2z}{z^2+\omega^2}$$ for $\omega\in\Bbb{C}$.   I'm pretty sure I know what happens when $\lvert \omega\rvert\neq 1$; it's simple: break it down into the two obvious cases. But what happens when $\lvert\omega\rvert =1$?    What I've done so far is let $z=\gamma (t)=e^{it}$ for $t\in [0, 2\pi ]$ and arrived at $$\frac{1}{\pi}\int_0^{2\pi}{\frac{dt}{1+\left(\frac{\omega}{e^{it}}\right)^2}},$$ hoping to use the derivative of $\operatorname{arctan}$, which isn't going to work, of course.  I'd like a few hints please :)  A hunch: I'm tempted to say that the integral does not exist for $\lvert\omega\rvert=1$ but I've no idea why.
42690
how to calculate $$\int_0^\infty \frac{x^{3}}{e^{x}-1} \; dx$$   Be $q:= e^{z}-1 , p:= z^{3}$ , then $e^{z} = 1 $ if $z= 2\pi n i $, so the residue at 0 is : $$\frac{p(z_{0})}{q&#39;(z_{0})} = 2\pi i n ^{3}$$  problem is that this is not symmetric, so how does one find the definite integral?
84554
What is $$\int\Gamma(x) \, dx$$ where $\Gamma(x)$ is the gamma function?
64695
Verify that $$\left|\int_{\gamma} \exp(iz^2)dz\right| \leq \frac{\pi\big(1-\exp(-r^2)\big)}{4r}$$ where $\gamma(t)=re^{it}$, for $0\leq t \leq \pi/4$ and $r &gt; 0$.   I'm stuck. here is my attempt:  $|\int_{\gamma} e^{i\cdot z^2}dz|\leq \int_{\gamma} |e^{i\cdot z^2}|dz=\int_{\gamma}|(e^{z^2})^i|dz$. As $t &gt; 0$ on $0\leq t \leq \pi/4$. $\Rightarrow|e^{r^2 e^{2it}}|=|e^{r^2}e^{e^{2it}}|&gt;0$ $\Rightarrow\int_{0}^{\pi/4} e^{r^2}e^{2it}rie^{it}dt=e^{r^2}ri\int_{0}^{\pi/4} exp(e^{2it}+it)dt$ Let $\alpha=e^{r^2}ri$ $\Rightarrow \alpha\int_{0}^{\pi/4}e^{\cos2t}e^{i\cdot \sin2t}(\cos(t)+i\cdot \sin(t)dt)$  Am I on track?
95905
I want to evaluate the integral $$\int_{\left | z \right |=3} \frac{e^{z}-1}{z(z-1)(z-i)}\,dz$$    by using the following  theorem:      If a function $f$ is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed curve $C$, then    $$\int_{C}f(z) dz=2\pi i\text{Res}\left ( \frac{1}{z^{2}}f(\frac{1}{z}), 0 \right ).$$   I want to calculate this integral  by calculating the residue at 0 of $\frac{1}{z^{2}}f(\frac{1}{z})=\frac{1+\frac{1}{2!z}+\frac{1}{3!z^{2}}\cdot \cdot \cdot }{1-(1+i)z+iz^{2}}$ So I did long division but the coefficient of 1/z is 0.  I know the method of calculating the residues at each singularities 0, 1, i, respectively. This method gives me the answer $2\pi i(0+\frac{e-1}{1-i}+\frac{e^{i}-1}{i(i-1)})$ which is not 0. Why answers are different?? Why does not hold long division method? Any help please.
61774
$\int_0^\infty \frac{1}{1+x^9} \, dx$  I tried taking the integral of $\Gamma_R = [0,R] \cup \gamma_R \cup I_R$, where we see that \gamma_R is the circle parametrized by $z = Re^{it}$ with $t\in[0,\frac{\pi}{2}]$. And $I_R$ is the line form $iR$ to $0$  one can note the following three things:  1) $\int_0^\infty \frac{1}{1+x^ 9}\,dz$ = $\int_0^\infty \frac{1}{1+z^ 9}\,dz$. (Note that $z$ is on the $x$-axis)  2) $\int_{\gamma_R}\frac{1}{1+z^ 9} = 0$. (By the ML inequality)  the main problem is when i parametrize $I^{-}_{R}$ with $z(x) = xi$ and $x \in [0,R]$. The main problem is that i get that:  3) $\int_{I_R}\frac{1}{1+z^9} = -\int_{I_R^-}\frac{1}{1+z^9} = -\int_{0}^{R} \frac{i}{ix^9 + 1}$.   note that i want to work towards taking $R$ to infinity eventually and then equal the sum of the integrals to the sum of the residuals of $\Gamma_R$ times $2\pi i$. The main problem here is that my term in 3), does not look like the form $c \cdot \int_{0}^{\infty}\frac{1}{1+x^ 9}$ with c a constant value. If i can get this done i think i'm finished.  PS: The singular points i got where $e^{\frac{\pi i}{9}},e^{\frac{3 \pi i}{9}}$.
12548
By considering the conjugate of its parametric form, evaluate $$\frac{1}{2\pi i}\int_{\gamma(0;1)}\frac{\overline{f(z)}}{z-a}dz$$ when $|a|&lt;1$ and $|a|&gt;1$, where $f$ is holomorphic in in the disk $(0;R),  R&gt;1$.  Typically when doing these kinds of integration and parametrization, $|z|=n$ is given, but it's different in this case (or is it not?). Can someone help me out?
36658
I know that $$ \int_0^\infty \frac{\log x}{\exp x} = -\gamma $$ where $ \gamma $ is the Euler-Mascheroni constant, but I have no idea how to prove this.  The series definition of $ \gamma $ leads me to believe that I should break the integrand into a series and interchange the summation and integration, but I can't think of a good series. The Maclaurin series of $ \ln x $ isn't applicable as the domain of $ x $ is not correct and I can't seem to manipulate the integrand so that such a Maclaurin series will work.   Another thing I thought of was using $ x \mapsto \log u $ to get $ \int\limits_{-\infty}^\infty \frac{\log \log u}{u^2} \ du $ and use some sort of contour integration, but I can't see how that would work out either.
54691
One way to evaluate $ \displaystyle\int_{0}^{z} \log \Gamma(x)  \,  \mathrm  dx $ is in terms of the .  $$ \int_{0}^{z} \log \Gamma(x)  \,  \mathrm  dx = \frac{z}{2}  \log (2 \pi) + \frac{z(1-z)}{2} +  z \log \Gamma(z) - \log G(z+1)$$  Another way is in terms of the .  $$ \int_{0}^{z} \log \Gamma(x)  \,  \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$  I've been trying to prove the latter so that I can prove $$\log G(z+1) - z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) .$$  My starting point is the  $$ \sum_{k=2}^{\infty} \zeta(k,a) x^{k-1} = \psi(a) - \psi(a-x) .$$  Integrating both sides, I get $$ \sum_{k=2}^{\infty} \frac{\zeta(k,a)}{k} x^{k} = \psi(a) x + \log \Gamma(a-x) - \log \Gamma(a),$$  which implies   $$ \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k,1)}{k} x^{k} = \gamma x + \log \Gamma(x+1) . $$   Then rearranging and integrating both sides from $0$ to $z$, I get  $$ \int_{0}^{z} \log \Gamma(x+1) \,  \mathrm dx = \int_{0}^{z} \log x \  \mathrm dx + \int_{0}^{z} \log \Gamma(x) \,  \mathrm dx   = - \frac{\gamma z^{2}}{2} + \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k,1)}{k(k+1)} z^{k+1} .$$  And then using the integral representation $$ \zeta(s,a) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-at}}{1-e^{-t}} \, \mathrm dt, $$ I get  $$ \int_{0}^{z} \log \Gamma(x) \,  \mathrm dx = z- z \log z  - \frac{\gamma z^{2}}{2} + \sum_{k=2}^{\infty} (-1)^{k} \frac{z^{k+1}}{k(k+1)} \frac{1}{\Gamma(k)} \int_{0}^{\infty} \frac{t^{k-1} e^{-t}}{1-e^{-t}} \, \mathrm dt $$  $$ = z - \log z - \frac{\gamma z^{2}}{2} + z \int_{0}^{\infty} \frac{e^{-t}}{1-e^{-t}} \frac{1}{t} \sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \frac{(zt)^{k}}{k!} \, \mathrm dt $$   $$ = z - z \log z - \frac{\gamma z^{2}}{2}  + z \int_{0}^{\infty} \frac{e^{-t}}{1-e^{-t}}\frac{1}{t} \left( -\frac{e^{-zt}}{zt} - 1 + \frac{zt}{2} + \frac{1}{zt} \right) \, \mathrm dt$$   $$ = z - z \log z - \frac{\gamma z^{2}}{2} + \lim_{s \to 0^{+}} \Big[ - \int_{0}^{\infty} \frac{t^{s -2} e^{-(z+1)t}}{1-e^{-t}} \, \mathrm dt - z \int_{0}^{\infty} \frac{t^{s-1} e^{-t}}{1-e^{-t}} \, \mathrm dt $$ $$ + \frac{z^{2}}{2} \int_{0}^{\infty} \frac{t^{s} e^{-t}}{1-e^{-t}} \, \mathrm dt  + \int_{0}^{\infty} \frac{t^{s -2} e^{-t}}{1-e^{-t}} \, \mathrm dt \Big]$$  $$ = z - z \log z - \frac{\gamma z^{2}}{2} + \lim_{s \to 0^{+}} \Big[ - \Gamma(s-1) \zeta(s-1,z+1) -z \Gamma(s) \zeta(s) + \frac{z^{2}}{2} \Gamma(s+1) \zeta(s+1)$$ $$ + \Gamma(s-1) \zeta(s-1) \Big] .$$  Assuming I haven't made any mistakes up to this point, how do I evaluate that limit?