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Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $

this isn't really a computable integral, But the only idea I have in mind is that:

derivative from $x=0$ to $x=1$ of: $\frac{d}{dx}\left({e^x}-{x})\right) = e -1 $, but I would still have to compute $\frac{e^x}{x^{2}+1}$,

because I'm not sure the derivatives will be sufficient in order to prove the inequality.

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    $\begingroup$ $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx < \int_0^1 e^x dx = e -1$ $\endgroup$
    – Martin R
    Feb 28, 2021 at 9:51
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    $\begingroup$ Use $x^2+1\geq 1$ $\endgroup$
    – Math-fun
    Feb 28, 2021 at 9:51

3 Answers 3

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Because of $\frac{1}{x^2+1}\leq 1$ for all $x$, we have $$\int_0^1 \frac{e^x}{x^2+1}dx\leq \int_0^1 e^x dx= e-1.$$

The inequality $\frac{1}{x^2+1}\leq 1$ follows from $x^2+1\geq 1$. This inequality is not very sharp though: The integrals equals, roughly, 1.271 whereas the upper bound is $e-1=1.718$.

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A better bound can be achieved by applying the Integral Chebyshev inequality to $f(x) = \frac{1}{1+x^2}$ and $g(x) = e^{x}$. Since $f$ and $g$ are of opposite monotonicity on $[0, 1]$, $$ \int_0^1 \frac{e^x}{1+x^2} \, dx \le \int_0^1 \frac{1}{1+x^2} \, dx \cdot \int_0^1 e^x \, dx = \frac{\pi}{4} (e-1) \approx 1.3495 \, . $$

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The most direct approach to this question is Triangle Inequality for Integrals. Below I provided a helpful link. https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals

Let us start by applying the inequality: $$\Biggl \lvert \int_{0}^{1} \frac{e^x}{x^2+1}dx\Biggl\rvert \space \leq \space \int_{0}^{1} \Biggl| \frac{e^x}{x^2+1}\Biggl|dx$$ Since we know that the function $\frac{e^x}{x^2+1}$ is positive-valued on the real line, we have: $$\int_{0}^{1} \Biggl| \frac{e^x}{x^2+1}\Biggl|dx = \int_{0}^{1} \frac{e^x}{x^2+1}dx$$ By the help of a simple algebraic fact $\frac{1}{1+x^2} \leq 1 \space \forall \space x \in \Bbb R$, it is easy to deduce that: $$\int_{0}^{1} \frac{e^x}{x^2+1}dx \space \leq \space \int_{0}^{1} e^x dx=[e^x]^{1}_{0}=e-1 $$

In conclusion: $$\bbox[yellow] {\int_{0}^{1} \frac{e^x}{x^2+1} dx \space \leq \space e-1 }$$

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  • $\begingroup$ What is the purpose of applying the triangle inequality if the integrand is positive? $\endgroup$
    – Martin R
    Mar 2, 2021 at 10:07

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