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This integral : $$\int_{-\infty}^{+\infty}\dfrac{\exp(-x^{2n+1}\arctan (x))}{1+x^{2n}}dx$$ seems converge for $n$ is a natural number according to some values of $n$ which I run in wolfram alpha , My simple justification is :since we have one of the known form $ \int u du $ such that $u=\arctan(x)$ occurs in our integral with the integral of $\exp$ function is $\exp$ function however we have the factor $x^{2n+1}$ in the side of arctan with odd exponent probably if we assume $f=\exp(-x^{2n+1}\arctan (x))$ then $ df=-(2n+1) x^{2n}\arctan (x))+\dfrac{- x^{2n+1}\arctan(x)}{1+x^2}$ , and if we use integration by parts and fraction however it would be complicated but we always have some luck to use $ \int u du $ , Now My question is : How do I justify the convergence of :$$\int_{-\infty}^{+\infty}\dfrac{\exp(-x^{2n+1}\arctan (x))}{1+x^{2n}}dx$$ with $n$ a positive integer?

clathratus
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zeraoulia rafik
  • 7,055

1 Answers1

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Observe that \begin{align*} 0<\frac{e^{-x^{2n}(x\arctan(x))}}{1+x^{2n}}\leq \frac{1}{1+x^{2n}} \end{align*} indeed for all $x\in \mathbb{R}$ we have $(x\arctan(x))\geq 0$ (remember that the function $\arctan(x)$ is odd).

Since $\int_\mathbb{R} \frac{1}{1+x^{2n}}dx$ converges for every $n\in \mathbb{N}^+$, by direct comparison test also our integral converges for every $n\in \mathbb{N}^+$.

The convergence of $\int_\mathbb{R} \frac{1}{1+x^{2n}}dx$ is follows easily by the convergence of $\int_1^\infty \frac{1}{x^2}dx$. Furthermore in this post the close formula for your integral is given.