Evaluate the value of: $$\int_0^1 e^{\left(x^2\right)}~dx+\int_1^e\sqrt{\ln x}~dx$$

Here is what I have tried:

$\ln x >\sqrt{\ln x}$ when $x=[1,e]$

$e^x>e^{\left(x^2\right)}$ when $x=[0,1]$

And after that put under the integral and obtain:

$$\int_0^1 e^{\left(x^2\right)}~dx+\int_1^e\sqrt{\ln x}~dx\leq e$$

And it is easy to see:

$$e\geq \int_0^1 e^{\left(x^2\right)}~dx + \int_1^e\sqrt{\ln x}~dx>1$$

But it is not solved just I know is smaller than $e$ and bigger than $1$

I can write a series but it will only give me an approximation.


Note that we can obtain a relationship between the two integrals via the substitution $x=e^{u^2}$ and integration by parts: $$\begin{align}\int_1^e \sqrt{\ln{x}}~dx&=\int_0^1 u\cdot 2ue^{u^2}~du\\&=\left[e^{u^2}u\right]_0^1-\int_0^1 e^{u^2}~du\\&=e-\int_0^1 e^{u^2}~du \tag{1}\end{align}$$ Hence, we deduce from $(1)$ that: $$\bbox[5px,border:2px solid #C0A000]{\int_0^1 e^{x^2}~dx+\int_1^e \sqrt{\ln{x}}~dx=e}$$

  • $\begingroup$ please explain me first row step by step $\endgroup$ – Ica Sandu Jul 12 '17 at 15:27
  • $\begingroup$ @IcaSandu: it is already explained: perform the substitution $x\mapsto e^{u^2}$, then apply integration by parts. $\endgroup$ – Jack D'Aurizio Jul 12 '17 at 15:37
  • $\begingroup$ in left you have integral from 1 to e and in right you have integral from 0 to 1 i dont understand how you change limit's of integral $\endgroup$ – Ica Sandu Jul 12 '17 at 15:40
  • $\begingroup$ i understand thank you ;) $\endgroup$ – Ica Sandu Jul 12 '17 at 15:52

HINT: If $f$ is an increasing function on the interval $[a,b]$ and $f^{-1}$ is its inverse function, then

$$ \int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\,f(b)-a\, f(a). \tag{1}$$


$\int\limits_0^1 e^{x^2}\ dx = \frac{1}{2} \sqrt{\pi } \text{erfi}(1)$


$\int\limits_1^e \sqrt{\ln (x)}\ dx = e-\frac{1}{2} \sqrt{\pi } \text{erfi}(1)$ (as projectilemotion showed)...

so the sum equals $e$.


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