Posts

On the closed form of the generalized square-sum arctan integral

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I was recently thinking about the closed-form expression to the $n$th order of the arctan integral in the case of the denominator of the integrated being a sum of squares. In context, the general sum of squares arctan integral is $$\int \dfrac{\mathrm{d}x}{x^2 + a^2} = \dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right) + C$$ $$\int_{0}^{\infty} \dfrac{\mathrm{d}x}{x^2 + a^2} = \dfrac{\pi}{2a}$$ A proof of this employs a basic substitution (factorization combined with $u = \frac{x}{a}$) and the noting of the arctan integral (a result that can be derived via implicit differentiate, with an example given on a previous post). Here, we shall consider the following integral, which we will evaluate using contour integration. $$\int_{\mathbb{R^{+}}} \dfrac{\mathrm{d}x}{x^n + a^n}$$ Factorizing $$\dfrac{1}{a^n} \int_{\mathbb{R^{+}}} \dfrac{\mathrm{d}x}{\left(\frac{x}{a}\right)^n + 1}$$ To simplify the denominator of our integrated, let $u = \frac{x}{a}$, with no change to our bounds, the reciprocal d

A rediscovery of Stirling's Approximation

I was adamant about trying to find my own way of computing non-integer values of the Gamma function. Extensions of the product definition of the Gamma function wouldn't directly work since those kinds of extensions would yield the same "accuracy" as the extensions of the zeta function to negative integer values. Hence, in doing this, I think I rediscovered a case of Stirling's approximation using a common technique that leverages logarithm laws. Firstly, consider the factorial function, we can see some examples below $$4! = (4)(3)(2)(1) = 24$$ $$7! = (7)(6)(5)(4)(3)(2)(1)$$ $$10! = (10)(9)(8)\cdots(1) = 3628800$$ and onwards, we can generalize this to $$n! = \prod_{\psi=1}^{n} (\psi)$$ Now, a common technique we use to deal with products, which I thought could be applied to this. Consider leveraging logarithm laws to express this finite product in terms of a series, the reason being that we have many more tools to approximate and evaluate series than we do finite prod

Taking an infinitely small bite

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We shall evaluate this contour integral today, given to me by a friend of mine. I hope to explore what it means when calculating residues (or seeing if we need to do that at all) at logarithmic poles. Our problem statement is the following  $$\int_{0}^{\infty}\frac{\ln{x}}{x^n + 1} dx$$ Examining poles at $$x = 0, \exp\left(\frac{i\pi}{n}\right),\exp\left(\frac{3i\pi}{n}\right) , \exp\left(\frac{5i\pi}{n}\right),\cdots,\exp\left(\frac{i(2k+1)\pi}{n}\right)$$ We will use this contour to isolate one pole, while not touching the pole at the origin $$\oint_C = \lim_{z \to e^{\frac{i\pi}{n}}} (z-e^\frac{i\pi}{n})\frac{\ln{z}}{z^n + 1} = \frac{2\pi^2}{n^2}e^{\frac{i\pi}{n}}$$ $$\oint_C = I+ \rho + \Omega = \frac{2\pi^2}{n^2}e^{\frac{i\pi}{n}}$$ $$\rho(z=Re^{iw}) = \int_{w=0}^{w=\frac{2\pi}{n}} \frac{Rie^{iw}\ln{R} - Rwe^{iw}}{R^ne^{iwn}+1}dw, R \to \infty, J \to 0$$ $$\Omega(z=te^{\frac{2\pi i}{n}}) = -\int_{\epsilon}^{R} e^\frac{2\pi i}{n} \cdot \left(\ln{t}+\frac{2\pi i}{n}\right)\cd

An interesting proof of the Binomial Theorem

In continuation of the previous Maclaurin series post, I present an interesting and intuitive proof of the binomial theorem. This proof presents and justifies the existence and dynamics of the binomial coefficient in quite a beautiful way.  The binomial coefficient is defined as the following $${n \choose r} = \dfrac{n!}{r! (n-r)!}$$ It acts as an analog to Pascal's triangle, with n being the row number and r is the column number. Consider the most basic binomial expansion $$(1+x)^n$$ We have chosen this as 1 acts trivially on the binomial coefficients. By the binomial theorem, we have $$(1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + \cdots + {n \choose n} x^n   $$ This resonates with the dynamics of a series expansion of the function, in hindsight it is, though it is finite. Let us explore the infinite version by means of taking the Maclaurin series of $(1+x)^n$. Letting $$f(x) = (1+x)^n$$ Then from the chain rule, it follows $$f'(x) = n(1+x)^{n-1}$$ $$f

A complex analog to the arctan function

 Trigonometric derivatives are fascinating, most specifically inverse trigonometric derivatives. I feel that they are the perfect example to showcase the true power of implicit differentiation. Consider the arctan function. Intuitively, it is difficult to compute any general form for its derivative, however we can do exactly this leveraging a key property of differentiation. Consider the following $$\arctan{x} = y(x)$$ $$x = \tan{[y(x)]}$$ Implicitly differentiating with respect $x$ on both side and applying the chain rule $$1 = y'(x) \cdot \sec^2{y(x)}$$ Noting the following $$y(x) = \arctan{x}$$ $$sec^2 {x} =1+ \tan^2{x}$$ Via substitution, we simplify our equation to  $$1 = y'(x) \cdot (\tan^2{[\arctan{x}}] + 1)$$ $$\dfrac{1}{x^2 + 1} = y'(x)$$ Which is the derivative of the arctan(x) function. Now, with about of curiosity, something quite interesting happens when we leverage this relation. Consider the following $$\int \dfrac{dx}{x^2 + 1} = \arctan{x}$$ Now, let us expa

On the derivation of the Maclaurin Series

I was researching and thinking about why there are factorials present in Taylor Series approximations, an interesting proof emerged, which I think suffices as a non-rigorous derivation of the Maclaurin series formula. Let us consider a continuous and differentiable function f(x). Now assume that we can approximate/express f(x) as an infinite order polynomial, mathematically $$f(x) = \sum_{i=0}^{\infty} a_{i}x^{i} = a_0 + a_1 x + a_2 x^2 \cdots$$ Now, let us assume we want to compute this approximation centered at $x=0$, in other words, find a Maclaurin series expression for f(x). Setting x to 0 $$f(0) = a_0 $$ Now, we note that every polynomial coefficient (a_i) will be directly correlated with the i-th derivative of f(x) centered at 0. This is justified as the derivative as a point by definition, tells us about the rate of change at that point, telling us about the shape of that function. Hence, let us consider $f'(x)$ $$f'(x) = \sum_{i=0}^{\infty} ia_{i}x^{i-1} = a_1 + 2a_2x 

MIT Integration Bee Qualifying Exam 2020

 In this post, we will explore the solutions to the first 5 integrals in the MIT integration bee. $$\int \dfrac{\log(2x)}{x\log(x)} dx$$ This can be solved using a simple substitution of $u=\log(x)$, hence the differential operator switch cancels out the x in the denominator and we are left with, also noting that $\log{x}$ and $\ln{x}$ are used interchangeably. $$\int \dfrac{u+\log{2}}{u} du$$ Expanding to $$\int 1 du + \dfrac{\log{2}}{u} du$$ which evaluates to $$ u + \log{2}\log{u} + C$$ recalling $u=\log(x)$, we find our solution to be $$\log(x) + \log(2)\log{(\log(x))} + C$$

A Simple MIT Integration Bee Integral

It has been a while since I have done an integral that does not involve Complex Analysis or Feynman substitutions, there we behold a hopefully beautiful MIT Integration Bee problem.  $$\int \dfrac{x+1}{x(x+\ln(x))} dx$$ To cancel out the x in the denominator, a clever u-sub would be $$u = \ln(x)$$ $$\int \dfrac{e^u+1}{u+e^u} du$$ We can see that the numerator is the derivative of the denominator, so $$\alpha = u + e^u$$ $$\int \dfrac{1}{\alpha} d\alpha$$ $$= \ln{\alpha} + C$$ $$= \ln({u+e^u}) + C$$ $$= \ln({\ln(x)+x}) + C$$ An elegant result and good practice of knowing which things to u-sub.

A MIT Engineering Contour Integral

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  This is what the inside of an advanced calculus course looks like for an MIT engineer! We will work through this long-answer problem as it guides us to the journey of evaluating an at-first daunting improper integral. $$\text{Source: MIT OCW, 2004, Advanced Calculus for Engineers}$$ Part 1 $$I = \text{Re} \left[\int_{-\infty}^{\infty} \dfrac{e^{iz}}{(4z^2 - \pi^2)(z^2+4)} dz\right]$$ We can compute our poles by solving the following equations $$(4z^2 - \pi^2)(z^2+4)=0$$ Getting poles at $$z = \pm \dfrac{\pi}{2}, z = \pm 2i$$ Part 2 We will be integrating along the upper half of the complex plane so we can leverage Jordan's Lemma. Our contour will look be Take a minute to admire the elegance & beauty of this contour, I drew it myself. Under the Residue Theorem $$\oint_C \dfrac{e^{iz}}{(4z^2 - \pi^2)(z^2+4)} dz = 2\pi i \sum{\text{Res}} f(z) = 2 \pi i \lim_{z \to 2i} \dfrac{(z-2i)(e^{iz})}{(4z^2 - \pi^2)(z^2+4)}$$ Applying L'Hôpital's rule once, we get $$ \lim_{z \to 2i

Sophomore's Dream | A Putnam experience

In my journey of practicing and tackling problems from the Putnam undergraduate exam, I came across what I know now as my favorite one.  $$\textbf{A Sophomore's Dream}$$ A beautiful identity first discovered by Johann Bernoulli, which I will now walk you through with my own take in proving this identity. The problem statement on the Putnam is $$\text{Prove the following}$$ $$\int_{0}^{1} \dfrac{dx}{x^x} = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} \cdots$$ An alternative definition of the RHS in terms of a series $$1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} \cdots = \sum_{n=1}^{\infty} \dfrac{1}{n^n}$$ Let us begin with the Taylor Series definition of the LHS integral above. Firstly noting that $$x^{-x} = e^{-x\ln{(x)}} = \alpha$$ Noting this to be $\alpha$ for ease of usage. We can use the $e^x$ Maclaurin series now to represent $\alpha$, Hence $$\sum_{n=0}^{\infty} \dfrac{(-x\ln{(x)})^n}{n!} = \alpha$$ Integrating term by term from 0 to 1 $$\int_{0}^{1} \alpha

Analytic continuation of the Gamma Function, A Cambridge Tripos Journey

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$$\textbf{The Euler Gamma function}$$ Usually denoted by $\Gamma(n)$, the Gamma function is an important result in analysis and number theory (due to its linkage with the zeta function). Using a clever trick, we can construct an analytic continuation  of this function (leveraging its integral definition) and reveal a lot more about its dynamics (poles, residues and etc.) in the complex plane. In this problem it is noted as $I(z)$ $${\textbf{Source: Cambridge MATHEMATICAL TRIPOS } \textit{Part II Tuesday, 8 September, 2020}}$$ (i) Show that $I(z+1) = zI(z)$ To do this, we can integrate $I(z)$ by parts $$I(z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt$$ Letting $ u = e^{-t} $ and $v' = t^{z-1}$, we get $$I(z) = \dfrac{1}{z} \int_{0}^{\infty} t^{z}e^{-t} dt$$ Simplifying under the definition of $I(z)$ $$I(z) = \dfrac{1}{z} I(z+1)$$ $$zI(z) = I(z+1)$$ $$\text{Hence this is shown}$$ (ii) Use part (I) to construct an analytic continuation of $I(z)$ into $\text{Re}$ $z > 0$, except at isol

A Cambridge Mathematical Tripos ODE problem

$$\textbf{Problem Statement}$$ $$\text{The function $y(x)$ satisfies the inhomogenous second-order linear differential equation }$$ $$y'' - 2y' - 3y = -16xe^{-x}$$ $$\text{Find the solution that satisfies the conditions that $y(0)=1$ and $y(x)$ is bounded as $x \to \infty$}$$ $$\textbf{Source : Cambridge Tripos 2001}$$ Upon first examination of this problem, there is a clear confusion on how to approach solving this, it is not a separable ODE and there are no boundary conditions for the first derivative so we cannot apply the Laplace Transform. We will combat this problem using a new & refreshing method called the Method of Undetermined Coefficients. It states that the solution to a differential equation can be given as $$y(x) = y_{p}(x) + y_{c}(x)$$ This will be explained later on, firstly we will try to find $$y_{p}(x)$$. To do this, we will guess the form of the answer and then convert this differential equations problem into a system of algebraic equations. Upon exa

A beautiful contour integral leveraging Jordan's Lemma

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 $$J = \int_{0}^{\infty} \dfrac{x^3 \sin(4x)}{x^4+4} dx = \text{Im} \left[\dfrac{1}{2} \oint_C \dfrac{z^3e^{4iz}}{z^4+4} dz\right]$$ We will integrate this counter-counterclockwise along the following contour We will do this since to find the value of our original integral, we would want to integrate on the upper half of the complex plane and with this we can also utilize Jordan's Lemma.  Knowing our original integral is even, we will integrate the complex function over from $-R$ to $R$ and then half our answer. We also utilize the complex definition of the sine and hence will need to take the imaginary part of our answer to get the solution to our integral. In our contour we have two main parts, one that runs along the real axis and a semi-circle $\gamma_R$, under Jordan's Lemma, this semi-circle yields to be 0, (there are also no poles on it). Now, integrating from $-R$ to $R$ gives us $2J$, hence we will simply use Cauchy's Residue theorem to calculate the residues of ou

An inductive proof of the general form of a Laplace Transform of a $n$th order derivative

We will prove the Laplace Transform of a $n$th order derivative is of the form of the following $$\mathcal{L} [ {f^{(n)}(x)}] =s^{n} \mathcal{L} [{f(x)}] - \sum_{i=1}^{n} s^{n-k}f^{(n-1)}(0)$$ We shall prove this for all $n \in \mathbb{N}$. In doing this we assume that $f(x)$ as like all of its derivatives are of exponential order. Induction attempt | Assume true for $n=k$, hence prove for $n=k+1$, Let $$f^{(k)}(x) = g(x)$$ Trivially $$f^{(k+1)}(x) = g'(x)$$ $$\mathcal{L} [g'(x)] = s \mathcal{L} [g(x)] - g(0)$$ Hence under substitution $$\mathcal{L} [f^{(k+1)}(x)] = s(s^{k} \mathcal{L} [{f(x)}] - \sum_{i=1}^{k} s^{k-i}f^{k-1}(0)) - f^{(k)}(0)$$ $$= s^{k+1} \mathcal{L} [{f(x)}] - \sum_{i=1}^{k} s^{k+1-i}f^{k}(0)$$ Simplifying to  $$= s^{k+1} \mathcal{L} [{f(x)}] - \sum_{i=1}^{k} s^{k-i}f^{k-1}(0)$$ This, under the principle of mathematical induction proves our initial claim for $n \in \mathbb{N}$.  $$\text{Quod Erat Demonstrandum}$$ This was quite a nice proof, maybe elementary

An elegant contour integral

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 $$\oint_C \dfrac{z^2}{z^4 + 1} dz$$ We want to integrate this over the upper half of the complex plane, with our original integral being $$I = \int_{-\infty}^{\infty} \dfrac{x^2}{x^4+1} dx$$ A picture of our contour Dividing our contour into two parts $J = -R$ to $R$ and $\gamma_R$ Firstly, examining the poles of our complex function, we see it has poles at $$z = \pm e^{\frac{i\pi}{4}}$$ $$z = \pm e^{\frac{3i\pi}{4}}$$ Integrating on our path J, we get I as R goes to infinity. We can also deduce that path $\gamma_R$ results in 0 (since it has no residues), the proof is an exercise left to the reader. Upon examination, we see that we have two poles in our contour $$z =  e^{\frac{i\pi}{4}}$$ $$z =  e^{\frac{3i\pi}{4}}$$ Hence under Cauchy's Residue thereom, our integral is simplified to the sum of the residues at these poles. We start at the first pole $$\lim _{z\to e^{\frac{i\pi }{4}}} \dfrac{(z-e^{\frac{i\pi }{4}})z^2}{z^4 + 1}$$ As this is in indeterminate form, we apply L'Hô

A long undergraduate TU Delft Differential Equation exam problem using Laplace Transforms

$$y'' + 2y' + 4y = \delta{(t-\pi)}$$ where $\delta$ denotes the Dirac Delta distribution. We have our given boundary conditions to be $y(0)=1$ and $y'(0) = 0$ Now, taking the Laplace Transform of both sides $$\mathcal{L} \left(y'' + 2y' + 4y \right)(s) = e^{-\pi s}$$ $$s^2 Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + 4Y(s) = e^{-\pi s}$$ Simplifying using our boundary conditions and factorizing $$(Y(s))(s^2+2s+4) = e^{-\pi s} + (s+2)$$ Rearranging to get $$Y(s) = \dfrac{e^{-\pi s}}{(s+1)^2 + \sqrt{3}^2} + \dfrac{s+1}{(s+1)^2 + \sqrt{3}^2} + \dfrac{\sqrt{3}}{(s+1)^2 + \sqrt{3}^2}$$ Let us denote these three fractions as variables $$ A = \dfrac{e^{-\pi s}}{(s+1)^2 + \sqrt{3}^2} $$ $$ B = \dfrac{s+1}{(s+1)^2 + \sqrt{3}^2 }$$ $$ C = \dfrac{\sqrt{3}}{(s+1)^2 + \sqrt{3}^2}$$ Using the Laplace Transform table provided, we take the inverse Laplace transform of each three with respect to x and find our function to be $$y(x) = e^{-x} \cos(\sqrt{3}t) + u(t-\pi)(e^{-t +

A neat solution to a beautiful integral with the help of ordinary differential equations.

$$\int_{-\infty}^{\infty} \dfrac{\cos{x}}{x^2+1} dx$$ Under symmetry, this is the same as $$2 \int_{0}^{\infty} \dfrac{\cos{x}}{x^2+1} dx$$ Under examination, we see that we will be required to use a clever technique known as the "Feynman Technique" or more mathematically recognized as the "Leibniz Rule" / "Differentiating under the integral sign". In doing so, we parameterize our integral $$I(a) = 2\int_{0}^{\infty} \dfrac{\cos{ax}}{x^2+1} dx$$  The derivative forms a special integral (the sine integral, we can include to be) $$I'(a) = -2\int_{0}^{\infty} \dfrac{a\sin{ax}}{x^2+1} dx - \pi$$  We can also note that $$I''(a) = I(a)$$  This form a differential equation, which we shall now solve. We can establish initial boundary conditions, knowing $\arctan{\infty} = \dfrac{\pi}{2}$. $$I(0) = \pi$$ $$I'(0) = -\pi$$ For simplicity, let $I(a) = y(x)$ $$y''=y$$ Noting the initial boundary conditions, through an ansatz we can deduce $$y =

A telescoping Putnam problem

 $$\int_{0}^{1} \dfrac{\ln{(x+1)}}{x^2+1} dx$$ Knowing $\tan^2{x} + 1 = \sec^2{x}$, we can use a trig-sub of $$x=\tan{\theta}$$ Hence  $$\dfrac{dx}{d\theta} = \sec^2{\theta}$$ Substituting this into our integral and simplifying bounds accordingly $$\int_{0}^{\frac{\pi}{4}} \dfrac{\ln{(\tan{\theta}+1)}}{\sec^2{\theta}} \cdot {\sec^2{\theta}} d\theta$$ $$I = \int_{0}^{\frac{\pi}{4}} \ln{(\tan{\theta}+1)} d\theta$$ Using the substitution $u = \frac{\pi}{4} - \theta$ $$I = -\int_{\frac{\pi}{4}}^{0} \ln\left({\tan{(\frac{\pi}{4}-u) +1}}\right) du$$ $$I = \int_{0}^{\frac{\pi}{4}} \ln\left({\tan{(\frac{\pi}{4}-u) +1}}\right) du$$ Expanding using the tangent addition identities $$I = \int_{0}^{\frac{\pi}{4}} \ln\left({\dfrac{2}{1+\tan{u}}}\right) du$$ Through examining the denominator and bounds, we can express the integral in terms of itself and everything beautifully cancels out, getting us $$2I = \int_{0}^{\frac{\pi}{4}} \ln{(2)} du$$ After evaluation, our answer to this problem would be  $

A IIT JEE Integral that reminds us of Euler's birthplace

 $$\int_{0}^{1} \dfrac{\ln(x)}{x-1} dx$$ We can bring in the substitution $u=x-1$ $$\int_{-1}^{0} \dfrac{\ln(u-1)}{u} du$$ We can recognize the numerator as a common Taylor Series, let us expand the integrand as such  $$\ln{(u+1)} = \sum_{n=1}^{\infty} \dfrac{(-u)^n}{n}$$ $$\dfrac{\ln{(u+1)}}{u} = \sum_{n=1}^{\infty} \dfrac{(-u)^{n-1}}{n}$$ Integrating both sides from -1 to 0 to get our simplified integral on the RHS $$\int_{-1}^{0} \dfrac{\ln{(u+1)}}{u} = \sum_{n=1}^{\infty} \dfrac{(1)^{n}}{n^2} $$ This is the famous Basel problem, named after Leonhard Euler's birthplace, the mathematician who first solved this problem. Henceforth our integral evaluates to $$\dfrac{\pi^2}{6}$$ This remained unsolved for hundreds of years, but in Euler's lifetime, he proposed not only one, but multiple proofs to the answer to this problem.

A Pleasant MIT Integration Bee substitution

$$\int{\dfrac{dx}{\sqrt{x}(1+\sqrt[3]{x})}}$$ Acknowledging the identity that $1+\tan^2{x} = \sec^2{x}$, we can introduce a substitution  $$x=\tan^{6}{\theta}$$ Hence $$\dfrac{dx}{d\theta} = 6\tan^{5}{\theta} \cdot sec^{2}{\theta}$$ Then our integral simplifies to $$6 \int \dfrac{\tan^{5}{\theta}sec^{2}{\theta}}{\tan^{3}{\theta}(\sec^{2}{\theta})} d\theta$$ Cancelling out like terms to $$6 \int \tan^{2}{\theta} d\theta$$ Using the identity discussed previously $$6 \int (\sec^{2}{\theta} - 1) d\theta$$ Using linearity of the integrals $$6 \int \sec^{2}{\theta}d\theta - 6 \int 1 d\theta$$ Knowing this is a common integral, we get $$6\tan{\theta} - 6\theta +C$$ Converting back to get our integrals in terms of x $$6(x^{\frac{1}{6}} - \arctan{(x^{\frac{1}{6}}))} +C $$

The King of Nested Radicals from the MIT Integration Bee

 $$\int{\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x}}}} \cdots} dx$$ This is a beautiful problem from the MIT Integration Bee. We can analyze the dynamics of this integrand and see it can be simplified, using exponent theory to  $$x^{\frac{1}{2}} \cdot x^{\frac{1}{2\cdot3}} \cdot x^{\frac{1}{2\cdot3\cdot4}} \cdot x^{\frac{1}{2\cdot3\cdot4\cdot5}} \cdots$$ Trivially $$x^{\frac{1}{2!}} \cdot x^{\frac{1}{3!}} \cdot x^{\frac{1}{4!}} \cdot x^{\frac{1}{5!}} \cdots x^{\frac{1}{n!}}$$ Upon examination, we can see this bears resemblance to the $e^{x}$ taylor series at $x=1$. Simplifying our integral to $$\int x^{e-2} dx$$ It follows that our integral evaluates to $$\dfrac{x^{e-1}}{{e-1}} + C$$ This is one of my favourite integrals, marrying in infinite nested radicals, Taylor series and integration all in one piece.

An Elegant solution to a Putnam Integral

$$\int_{0}^{\frac{\pi}{2}} \dfrac{dx}{1+(\tan{x})^{\sqrt{2}}}$$ We can use the substitution $$u = \frac{\pi}{2} - x$$ $$I =-\int_{\frac{\pi}{2}}^{0} \dfrac{dx}{1+(\cot{x})^{\sqrt{2}}}$$ Simplifying to $$I = \int_{0}^{\frac{\pi}{2}} \dfrac{dx}{1+(\cot{x})^{\sqrt{2}}}$$ Hence, now we can express the integral in terms of itself getting $$2I =  \int_{0}^{\frac{\pi}{2}} \dfrac{dx}{1+(\cot{x})^{\sqrt{2}}} +  \int_{0}^{\frac{\pi}{2}}  \dfrac{dx}{1+(\tan{x})^{\sqrt{2}}}$$ Using the Linearity of the Integral $$2I =  \int_{0}^{\frac{\pi}{2}} \dfrac{dx}{1+(\cot{x})^{\sqrt{2}}} \dfrac{dx}{1+(\tan{x})^{\sqrt{2}}}$$ Under simplification $$2I = \int_{0}^{\frac{\pi}{2}} \left(\dfrac{2+\cot(x)^{\sqrt{2}}+\tan(x)^{\sqrt{2}}}{2+\cot(x)^{\sqrt{2}}+\tan(x)^{\sqrt{2}}}\right) dx$$ $$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$$ $$2I = \dfrac{\pi}{2}$$ Simplifying to yield the solution to our integral, we get $$I = \dfrac{\pi}{4}$$