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How do we integrate the following two integrals? $$ \int_{-e}^\pi \cos(-3x^2) \, dx$$ and $$ \int_{-e}^\pi \frac {e^x}{\ln(a-x)} \, dx,$$ where $a>\pi$.

Souvik Dey
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    For the first one, take a look at http://math.stackexchange.com/questions/433448/how-do-i-integrate-int-01-sin-x2-dx/433453#433453 and http://en.wikipedia.org/wiki/Fresnel_integral – pitchounet Aug 26 '13 at 13:47
  • Maple by $J := int(cos(-3*x^2), x = -epsilon .. Pi)$ produces $$1/6,\sqrt {2}\sqrt {\pi }\sqrt {3}{\it FresnelC} \left( {\frac { \sqrt {2}\sqrt {3}\epsilon}{\sqrt {\pi }}} \right) +1/6,{\it FresnelC } \left( \sqrt {2}\sqrt {\pi }\sqrt {3} \right) \sqrt {2}\sqrt {\pi } \sqrt {3} $$ and $ series(J, epsilon, 3)$ gives $$1/6,{\it FresnelC} \left( \sqrt {2}\sqrt {\pi }\sqrt {3} \right) \sqrt {2}\sqrt {\pi }\sqrt {3}+\epsilon+O \left( {\epsilon}^{3} \right) .

    $$

     – user64494 Aug 26 '13 at 15:11
  • The Maple command $$series(int(exp(x)/ln(a-x), x = -epsilon .. Pi), epsilon, 2) assuming a > Pi $$ performs $$\int _{0}^{\pi }!{\frac {{{\rm e}^{x}}}{\ln \left( a-x \right) }}{dx }+{\frac {\epsilon}{\ln \left( a \right) }}+O \left( {\epsilon}^{2} \right) .$$ – user64494 Aug 26 '13 at 15:15

1 Answers1

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$\int_{-e}^\pi\cos(-3x^2)~dx$

$=\int_{-e}^\pi\cos3x^2~dx$

$=\int_{-e}^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^n(3x^2)^{2n}}{(2n)!}dx$

$=\int_{-e}^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^nx^{4n}}{(2n)!}dx$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^nx^{4n+1}}{(2n)!(4n+1)}\right]_{-e}^\pi$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^n(\pi^{4n+1}-(-e)^{4n+1})}{(2n)!(4n+1)}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^n(\pi^{4n+1}+e^{4n+1})}{(2n)!(4n+1)}$

$\int_{-e}^\pi\dfrac{e^x}{\ln(a-x)}dx$

$=\int_{a+e}^{a-\pi}\dfrac{e^{a-x}}{\ln x}d(a-x)$

$=\int_{a-\pi}^{a+e}\dfrac{e^ae^{-x}}{\ln x}dx$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^ae^{-e^x}}{x}d(e^x)$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^ae^xe^{-e^x}}{x}dx$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^ae^x}{x}\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^{nx}}{n!}dx$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^ae^{(n+1)x}}{n!x}dx$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^ne^a(n+1)^kx^{k-1}}{n!k!}dx$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^a}{n!x}dx+\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^kx^{k-1}}{n!k!}dx$

$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^{a-1}}{x}dx+\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^kx^{k-1}}{n!k!}dx$

$=\left[e^{a-1}\ln x\right]_{\ln(a-\pi)}^{\ln(a+e)}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^kx^k}{n!k!k}\right]_{\ln(a-\pi)}^{\ln(a+e)}$

$=e^{a-1}(\ln\ln(a+e)-\ln\ln(a-\pi))+\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^k((\ln(a+e))^k-(\ln(a-\pi))^k)}{n!k!k}$

Harry Peter
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