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Need to calculate this definite integral. It's seems very strange for me $$\int _0^{\frac{\pi }{2}}\sin \left(\arctan\left(x\right)+x\right)dx$$

I dont see any reasonable way to calculate this integral. For instance, arctan of π/2 - it's incomprehensible value. I think there are some clever and special way.

PersonaNonGrata
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    Question. Why would $\arctan{\frac{\pi}{2}}$ be an incomprehensible value? – Transcendental Sep 25 '15 at 18:38
  • PersonaNonGrata probably means that the arctan(pi/2) is not one of the standard trig values that everyone memorizes when they take analysis/calculus/trig. – Sinister Cutlass Sep 25 '15 at 18:39
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    I think the OP means that usually we de not see the domain of $\arctan$ as angles. However, $\arctan(\pi/2)$ is well defined ! – Alonso Delfín Sep 25 '15 at 18:40
  • PersonaNonGrata, are you contour integrals from complex analysis in your class? Maybe there's a way to reinterpret this integral in that setting. – Sinister Cutlass Sep 25 '15 at 18:41
  • Perhaps this is a substitution problem. Try u:=arctan(x), and rewrite dx in terms of du. Notice that the function becomes sin(u+tan(u)) and dx becomes ((tan(u))^2+1)du. – Sinister Cutlass Sep 25 '15 at 18:44
  • both maple and wolfram can't find it, even the definite integral – Jean-Sébastien Sep 25 '15 at 18:50
  • @Jean-Sébastien Really? Wolfram alpha tell me that is approx. 1.15727. – Asydot Sep 25 '15 at 18:57
  • @Asydot you are right, I guess I only tried the indefinite integral on wolfram. – Jean-Sébastien Sep 25 '15 at 18:59
  • i think there is no expression in the known elementary functions for this integral – Dr. Sonnhard Graubner Sep 25 '15 at 19:21
  • I mean if tan (π / 4) or tan (π / 6) is something "natural", the arctan of the same quantities are something incomprehensible. :)) Kind of like $\sin 3\sqrt2$: number makes sense but it is "artificial" as $3\sqrt2$ normally looks like sine value and not like argument. – PersonaNonGrata Sep 25 '15 at 19:41
  • @SinisterCutlass replacing $u = arctanx$ hardly something to give. We obtain the integral of $\frac{sin(u+tanu)}{1+u^2}$, and what to do with it? – PersonaNonGrata Sep 25 '15 at 19:43
  • @PersonaNonGrata if $u=arctan(x)$, then $du=\frac{1}{1+x^2} dx$, so $dx=(1+x^2)du=((1+(tan(u))^2)du$...... Now, $1+(tan(u))^2=(sec(u))^2$, and note that $\frac{d}{du} (tan(u)+u)=(sec(u))^2+1$. After this, maybe your integral can be attacked with ordinary substitution or integration by parts. – Sinister Cutlass Sep 25 '15 at 19:49

2 Answers2

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If the upper limit were $\infty$ instead of $\dfrac\pi2$ , then the result would be expressible in terms of Bessel I and Struve L functions. To prove this, first use the famous trigonometric formula for $\sin(a+b)$, then simplify $\sin(\arctan x)$ and $\cos(\arctan x)$, and rewrite $x\cos x+\sin x$ as the derivative of $x\sin x$, followed by integration by parts. Lastly, write $x^2=(x^2+1)-1$ to split up the integral into two nicer ones, and employ this to evaluate both of them. As it stands, however, the expression cannot be parsed even in terms of such special functions, unless, of course, one were to allow the existence of “incomplete” Bessel and Struve functions.

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Lucian
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0

$\int_0^\frac{\pi}{2}\sin(\tan^{-1}x+x)~dx$

$=\int_0^\frac{\pi}{2}\sin\tan^{-1}x\cos x~dx+\int_0^\frac{\pi}{2}\cos\tan^{-1}x\sin x~dx$

$=\int_0^\frac{\pi}{2}\dfrac{x\cos x}{\sqrt{x^2+1}}~dx+\int_0^\frac{\pi}{2}\dfrac{\sin x}{\sqrt{x^2+1}}~dx$

$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n)!\sqrt{x^2+1}}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!\sqrt{x^2+1}}~dx$

$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n)!\sqrt{x^2+1}}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!\sqrt{x^2+1}}~d(x^2+1)$

$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^n(-1)^{2n-k}(x^2+1)^k}{2(2n)!\sqrt{x^2+1}}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^n(-1)^{2n-k}(x^2+1)^k}{2(2n+1)!\sqrt{x^2+1}}~d(x^2+1)$

$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-\frac{1}{2}}}{2(2n)!k!(n-k)!}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-\frac{1}{2}}}{2(2n+1)!k!(n-k)!}~d(x^2+1)$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k+\frac{1}{2}}}{2(2n)!k!(n-k)!\left(k+\dfrac{1}{2}\right)}\right]_0^\frac{\pi}{2}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k+\frac{1}{2}}}{2(2n+1)!k!(n-k)!\left(k+\dfrac{1}{2}\right)}\right]_0^\frac{\pi}{2}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(\pi^2+2)^{k+\frac{1}{2}}}{2^{k+\frac{1}{2}}(2n)!k!(n-k)!(2k+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(\pi^2+2)^{k+\frac{1}{2}}}{2^{k+\frac{1}{2}}(2n+1)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n+1)!k!(n-k)!(2k+1)}$

Harry Peter
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