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

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.

• $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx < \int_0^1 e^x dx = e -1$ Feb 28, 2021 at 9:51
• Use $x^2+1\geq 1$ Feb 28, 2021 at 9:51

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 $$e-1=1.718$$.

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 $$[0, 1]$$, $$\int_0^1 \frac{e^x}{1+x^2} \, dx \le \int_0^1 \frac{1}{1+x^2} \, dx \cdot \int_0^1 e^x \, dx = \frac{\pi}{4} (e-1) \approx 1.3495 \, .$$

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

• What is the purpose of applying the triangle inequality if the integrand is positive? Mar 2, 2021 at 10:07