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If \( \large\displaystyle \int_{-\infty} ^{\infty} \left(\dfrac1e\right)^{x^2} \, dx = \sqrt{\pi} \), compute \( \Large \displaystyle \int_{-\infty}^{\infty} \dfrac{dx}{(e^x + 1) \sqrt[4]{e^{x^2} \sqrt{e^{2x^2} \sqrt{e^{3x^2} \cdots}}}} \).


[Source: All India Aakash Test Series]

Note by Akhilesh Prasad
1 year, 9 months ago

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\[\sqrt[4]{e^{x^2}\sqrt{e^{2x^2}.....}}=e^{1/2( x^2/2+2x^2/4+3x^2/8+.....)}=e^{x^2}\] the integral becomes \[\int_{-\infty}^\infty \dfrac{dx}{e^{x^2}(e^x+1)}\] now solve this.

Aareyan Manzoor - 1 year, 9 months ago

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@Aareyan Manzoor

Nihar Mahajan - 1 year, 9 months ago

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